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Transcript 
MATH 110 Sec 13.2 Operations with Events Practice Exercises
If the probability of surviving a head-on car accident at 55
mph is 0.067, what is the probability of not surviving?
MATH 110 Sec 13.2 Operations with Events Practice Exercises
If the probability of surviving a head-on car accident at 55
mph is 0.067, what is the probability of not surviving?
PROBABILITY OF THE
COMPLEMENT OF AN EVENT
If 𝐸 is an event, then 𝑃 𝐸 = 1 βˆ’ 𝑃(𝐸 β€² ).
MATH 110 Sec 13.2 Operations with Events Practice Exercises
If the probability of surviving a head-on car accident at 55
mph is 0.067, what is the probability of not surviving?
PROBABILITY OF THE
COMPLEMENT OF AN EVENT
If 𝐸 is an event, then 𝑃 𝐸 = 1 βˆ’ 𝑃(𝐸 β€² ).
𝑃 π‘›π‘œπ‘‘ π‘ π‘’π‘Ÿπ‘£π‘–π‘£π‘–π‘›π‘” = 1 βˆ’ 𝑃(π‘ π‘’π‘Ÿπ‘£π‘–π‘£π‘–π‘›π‘”)
MATH 110 Sec 13.2 Operations with Events Practice Exercises
If the probability of surviving a head-on car accident at 55
mph is 0.067, what is the probability of not surviving?
PROBABILITY OF THE
COMPLEMENT OF AN EVENT
If 𝐸 is an event, then 𝑃 𝐸 = 1 βˆ’ 𝑃(𝐸 β€² ).
𝑃 π‘›π‘œπ‘‘ π‘ π‘’π‘Ÿπ‘£π‘–π‘£π‘–π‘›π‘” = 1 βˆ’ 𝑃(π‘ π‘’π‘Ÿπ‘£π‘–π‘£π‘–π‘›π‘”)
𝑃 π‘›π‘œπ‘‘ π‘ π‘’π‘Ÿπ‘£π‘–π‘£π‘–π‘›π‘” = 1 βˆ’ 0.067
MATH 110 Sec 13.2 Operations with Events Practice Exercises
If the probability of surviving a head-on car accident at 55
mph is 0.067, what is the probability of not surviving?
PROBABILITY OF THE
COMPLEMENT OF AN EVENT
If 𝐸 is an event, then 𝑃 𝐸 = 1 βˆ’ 𝑃(𝐸 β€² ).
𝑃 π‘›π‘œπ‘‘ π‘ π‘’π‘Ÿπ‘£π‘–π‘£π‘–π‘›π‘” = 1 βˆ’ 𝑃(π‘ π‘’π‘Ÿπ‘£π‘–π‘£π‘–π‘›π‘”)
𝑃 π‘›π‘œπ‘‘ π‘ π‘’π‘Ÿπ‘£π‘–π‘£π‘–π‘›π‘” = 1 βˆ’ 0.067
𝑃 π‘›π‘œπ‘‘ π‘ π‘’π‘Ÿπ‘£π‘–π‘£π‘–π‘›π‘” = 0.933
MATH 110 Sec 13.2 Operations with Events Practice Exercises
A cafe has scratch-and-win tickets with a chance of winning a free
meal. If there is a 1 in 10 chance of winning a free meal:
What is the probability of winning a free meal?
MATH 110 Sec 13.2 Operations with Events Practice Exercises
A cafe has scratch-and-win tickets with a chance of winning a free
meal. If there is a 1 in 10 chance of winning a free meal:
Saying that there is a 1 in 10 chance is just another way to say that the
1
probability of winning a free meal is .
10
What is the probability of winning a free meal?
MATH 110 Sec 13.2 Operations with Events Practice Exercises
A cafe has scratch-and-win tickets with a chance of winning a free
meal. If there is a 1 in 10 chance of winning a free meal:
Saying that there is a 1 in 10 chance is just another way to say that the
1
probability of winning a free meal is .
10
What is the probability of winning a free meal?
1
𝑃 𝑀𝑖𝑛 π‘“π‘Ÿπ‘’π‘’ π‘šπ‘’π‘Žπ‘™ =
10
MATH 110 Sec 13.2 Operations with Events Practice Exercises
A cafe has scratch-and-win tickets with a chance of winning a free
meal. If there is a 1 in 10 chance of winning a free meal:
Saying that there is a 1 in 10 chance is just another way to say that the
1
probability of winning a free meal is .
10
What is the probability of winning a free meal?
1
𝑃 𝑀𝑖𝑛 π‘“π‘Ÿπ‘’π‘’ π‘šπ‘’π‘Žπ‘™ =
10
What is the probability that you do not win a free meal?
MATH 110 Sec 13.2 Operations with Events Practice Exercises
A cafe has scratch-and-win tickets with a chance of winning a free
meal. If there is a 1 in 10 chance of winning a free meal:
Saying that there is a 1 in 10 chance is just another way to say that the
1
probability of winning a free meal is .
10
What is the probability of winning a free meal?
1
𝑃 𝑀𝑖𝑛 π‘“π‘Ÿπ‘’π‘’ π‘šπ‘’π‘Žπ‘™ =
10
What is the probability that you do not win a free meal?
COMPLEMENT RULE: 𝑃 𝐸 = 1 βˆ’ 𝑃(𝐸 β€² )
MATH 110 Sec 13.2 Operations with Events Practice Exercises
A cafe has scratch-and-win tickets with a chance of winning a free
meal. If there is a 1 in 10 chance of winning a free meal:
Saying that there is a 1 in 10 chance is just another way to say that the
1
probability of winning a free meal is .
10
What is the probability of winning a free meal?
1
𝑃 𝑀𝑖𝑛 π‘“π‘Ÿπ‘’π‘’ π‘šπ‘’π‘Žπ‘™ =
10
What is the probability that you do not win a free meal?
COMPLEMENT RULE: 𝑃 𝐸 = 1 βˆ’ 𝑃(𝐸 β€² )
𝑃
π‘‘π‘œπ‘›β€² 𝑑
𝑀𝑖𝑛 = 1 βˆ’ 𝑃 𝑀𝑖𝑛 π‘“π‘Ÿπ‘’π‘’ π‘šπ‘’π‘Žπ‘™ = 1
1
βˆ’
10
=
9
10
MATH 110 Sec 13.2 Operations with Events Practice Exercises
A cafe has scratch-and-win tickets with a chance of winning a free
meal. If there is a 1 in 10 chance of winning a free meal:
Saying that there is a 1 in 10 chance is just another way to say that the
1
probability of winning a free meal is .
10
What is the probability of winning a free meal?
1
𝑃 𝑀𝑖𝑛 π‘“π‘Ÿπ‘’π‘’ π‘šπ‘’π‘Žπ‘™ =
10
What is the probability that you do not win a free meal?
COMPLEMENT RULE: 𝑃 𝐸 = 1 βˆ’ 𝑃(𝐸 β€² )
𝑃
π‘‘π‘œπ‘›β€² 𝑑
𝑀𝑖𝑛 = 1 βˆ’ 𝑃 𝑀𝑖𝑛 π‘“π‘Ÿπ‘’π‘’ π‘šπ‘’π‘Žπ‘™ = 1
1
βˆ’
10
=
9
10
MATH 110 Sec 13.2 Operations with Events Practice Exercises
A cafe has scratch-and-win tickets with a chance of winning a free
meal. If there is a 1 in 10 chance of winning a free meal:
Saying that there is a 1 in 10 chance is just another way to say that the
1
probability of winning a free meal is .
10
What is the probability of winning a free meal?
1
𝑃 𝑀𝑖𝑛 π‘“π‘Ÿπ‘’π‘’ π‘šπ‘’π‘Žπ‘™ =
10
What is the probability that you do not win a free meal?
COMPLEMENT RULE: 𝑃 𝐸 = 1 βˆ’ 𝑃(𝐸 β€² )
𝑃
π‘‘π‘œπ‘›β€² 𝑑
𝑀𝑖𝑛 = 1 βˆ’ 𝑃 𝑀𝑖𝑛 π‘“π‘Ÿπ‘’π‘’ π‘šπ‘’π‘Žπ‘™ = 1
1
βˆ’
10
=
9
10
MATH 110 Sec 13.2 Operations with Events Practice Exercises
If 10 coins are flipped, what is the
probability of obtaining at least one head?
MATH 110 Sec 13.2 Operations with Events Practice Exercises
If 10 coins are flipped, what is the
probability of obtaining at least one head?
We are interested in the number of heads so visualize this:
MATH 110 Sec 13.2 Operations with Events Practice Exercises
If 10 coins are flipped, what is the
probability of obtaining at least one head?
We are interested in the number of heads so visualize this:
0
1
2
Possible Number of Heads
3
4
5
6
7
8
9
10
MATH 110 Sec 13.2 Operations with Events Practice Exercises
If 10 coins are flipped, what is the
probability of obtaining at least one head?
We are interested in the number of heads so visualize this:
0
1
2
Possible Number of Heads
3
4
5
6
7
8
9
10
β€˜At least one head’ means one of these 9 possibilities
MATH 110 Sec 13.2 Operations with Events Practice Exercises
If 10 coins are flipped, what is the
probability of obtaining at least one head?
We are interested in the number of heads so visualize this:
0
1
2
Possible Number of Heads
3
4
5
6
7
8
9
10
β€˜At least one head’ means one of these 9 possibilities
This makes it clear that it is MUCH easier to find 𝑃(0 β„Žπ‘’π‘Žπ‘‘π‘ )
and then use the Complement Rule.
MATH 110 Sec 13.2 Operations with Events Practice Exercises
If 10 coins are flipped, what is the
probability of obtaining at least one head?
We are interested in the number of heads so visualize this:
0
1
2
Possible Number of Heads
3
4
5
6
7
8
9
10
β€˜At least one head’ means one of these 9 possibilities
This makes it clear that it is MUCH easier to find 𝑃(0 β„Žπ‘’π‘Žπ‘‘π‘ )
and then use the Complement Rule.
𝑃 π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ π‘œπ‘›π‘’ β„Žπ‘’π‘Žπ‘‘ = 1 βˆ’ 𝑃 0 β„Žπ‘’π‘Žπ‘‘π‘  =
9
10
MATH 110 Sec 13.2 Operations with Events Practice Exercises
If 10 coins are flipped, what is the
probability of obtaining at least one head?
We are interested
the
number
heads. so visualize this:
So, in
let’s
calculate
𝑃 0ofβ„Žπ‘’π‘Žπ‘‘π‘ 
Because on eachPossible
flip, we Number
get eitherofHHeads
or T (2 possibilities),
Principle
there
are
0 the
1 Fundamental
2
3 Counting
4
5
6 says7that
8
9
10
10
2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2 = 1024
different
that
10 coins
canofbethese
flipped.
β€˜At least
oneways
head’
means
one
9 possibilities
BUT only 1 of these 1024 ways (TTTTTTTTTT) is β€˜0 heads’.
1
This makes it clear that
MUCH=easier
So, 𝑃it0isβ„Žπ‘’π‘Žπ‘‘π‘ 
. to find 𝑃(0 β„Žπ‘’π‘Žπ‘‘π‘ )
1024
and then use the Complement
Rule.
𝑃 π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ π‘œπ‘›π‘’ β„Žπ‘’π‘Žπ‘‘ = 1 βˆ’ 𝑃 0 β„Žπ‘’π‘Žπ‘‘π‘  =
9
10
MATH 110 Sec 13.2 Operations with Events Practice Exercises
If 10 coins are flipped, what is the
probability of obtaining at least one head?
We are interested
the
number
heads. so visualize this:
So, in
let’s
calculate
𝑃 0ofβ„Žπ‘’π‘Žπ‘‘π‘ 
Because on eachPossible
flip, we Number
get eitherofHHeads
or T (2 possibilities),
Principle
there
are
0 the
1 Fundamental
2
3 Counting
4
5
6 says7that
8
9
10
10
2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2 = 1024
different
that
10 coins
canofbethese
flipped.
β€˜At least
oneways
head’
means
one
9 possibilities
BUT only 1 of these 1024 ways (TTTTTTTTTT) is β€˜0 heads’.
1
This makes it clear that
MUCH=easier
So, 𝑃it0isβ„Žπ‘’π‘Žπ‘‘π‘ 
. to find 𝑃(0 β„Žπ‘’π‘Žπ‘‘π‘ )
1024
and then use the Complement
Rule.
𝑃 π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ π‘œπ‘›π‘’ β„Žπ‘’π‘Žπ‘‘ = 1 βˆ’ 𝑃 0 β„Žπ‘’π‘Žπ‘‘π‘  =
9
10
MATH 110 Sec 13.2 Operations with Events Practice Exercises
If 10 coins are flipped, what is the
probability of obtaining at least one head?
We are interested
the
number
heads. so visualize this:
So, in
let’s
calculate
𝑃 0ofβ„Žπ‘’π‘Žπ‘‘π‘ 
Because on eachPossible
flip, we Number
get eitherofHHeads
or T (2 possibilities),
Principle
there
are
0 the
1 Fundamental
2
3 Counting
4
5
6 says7that
8
9
10
10
2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2 = 1024
different
that
10 coins
canofbethese
flipped.
β€˜At least
oneways
head’
means
one
9 possibilities
BUT only 1 of these 1024 ways (TTTTTTTTTT) is β€˜0 heads’.
1
This makes it clear that
MUCH=easier
So, 𝑃it0isβ„Žπ‘’π‘Žπ‘‘π‘ 
. to find 𝑃(0 β„Žπ‘’π‘Žπ‘‘π‘ )
1024
and then use the Complement
Rule.
𝑃 π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ π‘œπ‘›π‘’ β„Žπ‘’π‘Žπ‘‘ = 1 βˆ’ 𝑃 0 β„Žπ‘’π‘Žπ‘‘π‘  =
9
10
MATH 110 Sec 13.2 Operations with Events Practice Exercises
If 10 coins are flipped, what is the
probability of obtaining at least one head?
We are interested
the
number
heads. so visualize this:
So, in
let’s
calculate
𝑃 0ofβ„Žπ‘’π‘Žπ‘‘π‘ 
Because on eachPossible
flip, we Number
get eitherofHHeads
or T (2 possibilities),
Principle
there
are
0 the
1 Fundamental
2
3 Counting
4
5
6 says7that
8
9
10
10
2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2 = 1024
different
that
10 coins
canofbethese
flipped.
β€˜At least
oneways
head’
means
one
9 possibilities
BUT only 1 of these 1024 ways (TTTTTTTTTT) is β€˜0 heads’.
1
This makes it clear that
MUCH=easier
So, 𝑃it0isβ„Žπ‘’π‘Žπ‘‘π‘ 
. to find 𝑃(0 β„Žπ‘’π‘Žπ‘‘π‘ )
1024
and then use the Complement
Rule.
𝑃 π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ π‘œπ‘›π‘’ β„Žπ‘’π‘Žπ‘‘ = 1 βˆ’ 𝑃 0 β„Žπ‘’π‘Žπ‘‘π‘  =
9
10
MATH 110 Sec 13.2 Operations with Events Practice Exercises
If 10 coins are flipped, what is the
probability of obtaining at least one head?
We are interested
the
number
heads. so visualize this:
So, in
let’s
calculate
𝑃 0ofβ„Žπ‘’π‘Žπ‘‘π‘ 
Because on eachPossible
flip, we Number
get eitherofHHeads
or T (2 possibilities),
Principle
there
are
0 the
1 Fundamental
2
3 Counting
4
5
6 says7that
8
9
10
10
2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2 = 1024
different
that
10 coins
canofbethese
flipped.
β€˜At least
oneways
head’
means
one
9 possibilities
BUT only 1 of these 1024 ways (TTTTTTTTTT) is β€˜0 heads’.
1
This makes it clear that
MUCH=easier
So, 𝑃it0isβ„Žπ‘’π‘Žπ‘‘π‘ 
. to find 𝑃(0 β„Žπ‘’π‘Žπ‘‘π‘ )
1024
and then use the Complement
Rule.
𝑃 π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ π‘œπ‘›π‘’ β„Žπ‘’π‘Žπ‘‘ = 1 βˆ’ 𝑃 0 β„Žπ‘’π‘Žπ‘‘π‘  =
9
10
MATH 110 Sec 13.2 Operations with Events Practice Exercises
If 10 coins are flipped, what is the
probability of obtaining at least one head?
We are interested
the
number
heads. so visualize this:
So, in
let’s
calculate
𝑃 0ofβ„Žπ‘’π‘Žπ‘‘π‘ 
Because on eachPossible
flip, we Number
get eitherofHHeads
or T (2 possibilities),
Principle
there
are
0 the
1 Fundamental
2
3 Counting
4
5
6 says7that
8
9
10
10
2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2 = 1024
different
that
10 coins
canofbethese
flipped.
β€˜At least
oneways
head’
means
one
9 possibilities
BUT only 1 of these 1024 ways (TTTTTTTTTT) is β€˜0 heads’.
1
This makes it clear that
MUCH=easier
So, 𝑃it0isβ„Žπ‘’π‘Žπ‘‘π‘ 
. to find 𝑃(0 β„Žπ‘’π‘Žπ‘‘π‘ )
1024
and then use the Complement
Rule.
𝑃 π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ π‘œπ‘›π‘’ β„Žπ‘’π‘Žπ‘‘ = 1 βˆ’ 𝑃 0 β„Žπ‘’π‘Žπ‘‘π‘  =
9
10
MATH 110 Sec 13.2 Operations with Events Practice Exercises
If 10 coins are flipped, what is the
probability of obtaining at least one head?
We are interested
the
number
heads. so visualize this:
So, in
let’s
calculate
𝑃 0ofβ„Žπ‘’π‘Žπ‘‘π‘ 
Because on eachPossible
flip, we Number
get eitherofHHeads
or T (2 possibilities),
Principle
there
are
0 the
1 Fundamental
2
3 Counting
4
5
6 says7that
8
9
10
10
2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2 = 1024
different
that
10 coins
canofbethese
flipped.
β€˜At least
oneways
head’
means
one
9 possibilities
BUT only 1 of these 1024 ways (TTTTTTTTTT) is β€˜0 heads’.
1
This makes it clear that
MUCH=easier
So, 𝑃it0isβ„Žπ‘’π‘Žπ‘‘π‘ 
. to find 𝑃(0 β„Žπ‘’π‘Žπ‘‘π‘ )
1024
and then use the Complement
Rule.
1
1023 9
=
𝑃 π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ π‘œπ‘›π‘’ β„Žπ‘’π‘Žπ‘‘ = 1 βˆ’ 𝑃 0 β„Žπ‘’π‘Žπ‘‘π‘ 
=
1024 1024 10
MATH 110 Sec 13.2 Operations with Events Practice Exercises
If 10 coins are flipped, what is the
probability of obtaining at least one head?
We are interested
the
number
heads. so visualize this:
So, in
let’s
calculate
𝑃 0ofβ„Žπ‘’π‘Žπ‘‘π‘ 
Because on eachPossible
flip, we Number
get eitherofHHeads
or T (2 possibilities),
Principle
there
are
0 the
1 Fundamental
2
3 Counting
4
5
6 says7that
8
9
10
10
2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2 = 1024
different
that
10 coins
canofbethese
flipped.
β€˜At least
oneways
head’
means
one
9 possibilities
BUT only 1 of these 1024 ways (TTTTTTTTTT) is β€˜0 heads’.
1
This makes it clear that
MUCH=easier
So, 𝑃it0isβ„Žπ‘’π‘Žπ‘‘π‘ 
. to find 𝑃(0 β„Žπ‘’π‘Žπ‘‘π‘ )
1024
and then use the Complement
Rule.
1
1023 9
=
𝑃 π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ π‘œπ‘›π‘’ β„Žπ‘’π‘Žπ‘‘ = 1 βˆ’ 𝑃 0 β„Žπ‘’π‘Žπ‘‘π‘ 
=
1024 1024 10
MATH 110 Sec 13.2 Operations with Events Practice Exercises
If 10 coins are flipped, what is the
probability of obtaining at least one head?
We are interested in the number of heads so visualize this:
0
1
2
Possible Number of Heads
3
4
5
6
7
8
9
10
This makes it clear that it is MUCH easier to find 𝑃(0 β„Žπ‘’π‘Žπ‘‘π‘ )
and then use the Complement Rule.
1
1023 9
=
𝑃 π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ π‘œπ‘›π‘’ β„Žπ‘’π‘Žπ‘‘ = 1 βˆ’ 𝑃 0 β„Žπ‘’π‘Žπ‘‘π‘ 
=
1024 1024 10
MATH 110 Sec 13.2 Operations with Events Practice Exercises
What is the probability of getting either a heart or a queen
when drawing a single card from a deck of 52 cards?
MATH 110 SecStandard
13-2 Lecture:
Operations
With Events
deck of
cards
4 suits (CLUBS, SPADES, HEARTS, DIAMONDS)
4 Queens
If we select a single card from a standard 52-card deck, what
is the probability that we draw either a heart or a face card?
Let H be the event β€˜draw a heart’ and F be β€˜draw a face card’.
That means that we are looking for 𝑃(𝐻 βˆͺ 𝐹).
13 HEARTS
Recall what a standard 52-card deck looks like.
MATH 110 SecStandard
13-2 Lecture:
Operations
With Events
deck of
cards
4 suits (CLUBS, SPADES, HEARTS, DIAMONDS)
4 Queens
If we select a single card from a standard 52-card deck, what
is the probability that we draw either a heart or a face card?
Let H be theThere
eventare
β€˜draw
a heart’
F be β€˜draw a face card’.
13 Hearts
and 4and
Queens
BUT thethat
Queen
Hearts
gets counted
That means
weofare
looking
for 𝑃(𝐻twice.
βˆͺ 𝐹).
So, the number of Hearts or Queens is 13 + 4 – 1 = 16.
13 HEARTS
Recall what a standard 52-card deck looks like.
MATH 110 SecStandard
13-2 Lecture:
Operations
With Events
deck of
cards
4 suits (CLUBS, SPADES, HEARTS, DIAMONDS)
4 Queens
If we select a single card from a standard 52-card deck, what
is the probability that we draw either a heart or a face card?
Let H be theThere
eventare
β€˜draw
a heart’
F be β€˜draw a face card’.
13 Hearts
and 4and
Queens
BUT thethat
Queen
Hearts
gets counted
That means
weofare
looking
for 𝑃(𝐻twice.
βˆͺ 𝐹).
So, the number of Hearts or Queens is 13 + 4 – 1 = 16.
13 HEARTS
Recall what a standard 52-card deck looks like.
MATH 110 SecStandard
13-2 Lecture:
Operations
With Events
deck of
cards
4 suits (CLUBS, SPADES, HEARTS, DIAMONDS)
4 Queens
If we select a single card from a standard 52-card deck, what
is the probability that we draw either a heart or a face card?
Let H be theThere
eventare
β€˜draw
a heart’
F be β€˜draw a face card’.
13 Hearts
and 4and
Queens
BUT thethat
Queen
Hearts
gets counted
That means
weofare
looking
for 𝑃(𝐻twice.
βˆͺ 𝐹).
So, the number of Hearts or Queens is 13 + 4 – 1 = 16.
13 HEARTS
Recall what a standard 52-card deck looks like.
MATH 110 Sec 13.2 Operations with Events Practice Exercises
What is the probability of getting either a heart or a queen
when drawing a single card from a deck of 52 cards?
There are 13 Hearts and 4 Queens
BUT the Queen of Hearts gets counted twice.
So, the number of Hearts or Queens is 13 + 4 – 1 = 16.
MATH 110 Sec 13.2 Operations with Events Practice Exercises
What is the probability of getting either a heart or a queen
when drawing a single card from a deck of 52 cards?
There are 13 Hearts and 4 Queens
BUT the Queen of Hearts gets counted twice.
So, the number of Hearts or Queens is 13 + 4 – 1 = 16.
By the Basic Probability Principle
# π‘œπ‘“ π»π‘’π‘Žπ‘Ÿπ‘‘π‘  π‘œπ‘Ÿ 𝑄𝑒𝑒𝑒𝑛𝑠 16
4
𝑃 π»π‘’π‘Žπ‘Ÿπ‘‘ π‘œπ‘Ÿ 𝑄𝑒𝑒𝑒𝑛 =
=
=
# π‘œπ‘“ π‘π‘Žπ‘Ÿπ‘‘π‘ 
52 13
MATH 110 Sec 13.2 Operations with Events Practice Exercises
What is the probability of getting either a heart or a queen
when drawing a single card from a deck of 52 cards?
There are 13 Hearts and 4 Queens
BUT the Queen of Hearts gets counted twice.
So, the number of Hearts or Queens is 13 + 4 – 1 = 16.
By the Basic Probability Principle
# π‘œπ‘“ π»π‘’π‘Žπ‘Ÿπ‘‘π‘  π‘œπ‘Ÿ 𝑄𝑒𝑒𝑒𝑛𝑠 16
4
𝑃 π»π‘’π‘Žπ‘Ÿπ‘‘ π‘œπ‘Ÿ 𝑄𝑒𝑒𝑒𝑛 =
=
=
# π‘œπ‘“ π‘π‘Žπ‘Ÿπ‘‘π‘ 
52 13
MATH 110 Sec 13.2 Operations with Events Practice Exercises
What is the probability of getting either a heart or a queen
when drawing a single card from a deck of 52 cards?
There are 13 Hearts and 4 Queens
BUT the Queen of Hearts gets counted twice.
So, the number of Hearts or Queens is 13 + 4 – 1 = 16.
By the Basic Probability Principle
# π‘œπ‘“ π»π‘’π‘Žπ‘Ÿπ‘‘π‘  π‘œπ‘Ÿ 𝑄𝑒𝑒𝑒𝑛𝑠 16
4
𝑃 π»π‘’π‘Žπ‘Ÿπ‘‘ π‘œπ‘Ÿ 𝑄𝑒𝑒𝑒𝑛 =
=
=
# π‘œπ‘“ π‘π‘Žπ‘Ÿπ‘‘π‘ 
52 13
MATH 110 Sec 13.2 Operations with Events Practice Exercises
What is the probability of getting either a heart or a queen
when drawing a single card from a deck of 52 cards?
There are 13 Hearts and 4 Queens
BUT the Queen of Hearts gets counted twice.
So, the number of Hearts or Queens is 13 + 4 – 1 = 16.
By the Basic Probability Principle
# π‘œπ‘“ π»π‘’π‘Žπ‘Ÿπ‘‘π‘  π‘œπ‘Ÿ 𝑄𝑒𝑒𝑒𝑛𝑠 16
4
𝑃 π»π‘’π‘Žπ‘Ÿπ‘‘ π‘œπ‘Ÿ 𝑄𝑒𝑒𝑒𝑛 =
=
=
# π‘œπ‘“ π‘π‘Žπ‘Ÿπ‘‘π‘ 
52 13
MATH 110 Sec 13.2 Operations with Events Practice Exercises
Assume A and B are events.
If P(A U B) = 0.7, P(A) = 0.3 and P(B) = 0.65, find P(A ∩ B).
MATH 110 Sec 13.2 Operations with Events Practice Exercises
Assume A and B are events.
If P(A U B) = 0.7, P(A) = 0.3 and P(B) = 0.65, find P(A ∩ B).
UNION RULE FOR PROBABILITIES
𝑃 𝐴 βˆͺ 𝐡 = 𝑃 𝐴 + 𝑃 𝐡 βˆ’ 𝑃(𝐴 ∩ 𝐡)
MATH 110 Sec 13.2 Operations with Events Practice Exercises
Assume A and B are events.
If P(A U B) = 0.7, P(A) = 0.3 and P(B) = 0.65, find P(A ∩ B).
UNION RULE FOR PROBABILITIES
𝑃 𝐴 βˆͺ 𝐡 = 𝑃 𝐴 + 𝑃 𝐡 βˆ’ 𝑃(𝐴 ∩ 𝐡)
MATH 110 Sec 13.2 Operations with Events Practice Exercises
Assume A and B are events.
If P(A U B) = 0.7, P(A) = 0.3 and P(B) = 0.65, find P(A ∩ B).
UNION RULE FOR PROBABILITIES
𝑃 𝐴 βˆͺ 𝐡 = 𝑃 𝐴 + 𝑃 𝐡 βˆ’ 𝑃(𝐴 ∩ 𝐡)
0.7 = 0.3 + 0.65 βˆ’ 𝑃(𝐴 ∩ 𝐡)
MATH 110 Sec 13.2 Operations with Events Practice Exercises
Assume A and B are events.
If P(A U B) = 0.7, P(A) = 0.3 and P(B) = 0.65, find P(A ∩ B).
UNION RULE FOR PROBABILITIES
𝑃 𝐴 βˆͺ 𝐡 = 𝑃 𝐴 + 𝑃 𝐡 βˆ’ 𝑃(𝐴 ∩ 𝐡)
0.7 = 0.3 + 0.65 βˆ’ 𝑃(𝐴 ∩ 𝐡)
MATH 110 Sec 13.2 Operations with Events Practice Exercises
Assume A and B are events.
If P(A U B) = 0.7, P(A) = 0.3 and P(B) = 0.65, find P(A ∩ B).
UNION RULE FOR PROBABILITIES
𝑃 𝐴 βˆͺ 𝐡 = 𝑃 𝐴 + 𝑃 𝐡 βˆ’ 𝑃(𝐴 ∩ 𝐡)
0.7 = 0.3 + 0.65 βˆ’ 𝑃(𝐴 ∩ 𝐡)
0.7 = 0.95 βˆ’ 𝑃(𝐴 ∩ 𝐡)
MATH 110 Sec 13.2 Operations with Events Practice Exercises
Assume A and B are events.
If P(A U B) = 0.7, P(A) = 0.3 and P(B) = 0.65, find P(A ∩ B).
UNION RULE FOR PROBABILITIES
𝑃 𝐴 βˆͺ 𝐡 = 𝑃 𝐴 + 𝑃 𝐡 βˆ’ 𝑃(𝐴 ∩ 𝐡)
0.7 = 0.3 + 0.65 βˆ’ 𝑃(𝐴 ∩ 𝐡)
0.7 = 0.95 βˆ’ 𝑃(𝐴 ∩ 𝐡)
𝑃(𝐴 ∩ 𝐡) = 0.95 βˆ’ 0.70
MATH 110 Sec 13.2 Operations with Events Practice Exercises
Assume A and B are events.
If P(A U B) = 0.7, P(A) = 0.3 and P(B) = 0.65, find P(A ∩ B).
UNION RULE FOR PROBABILITIES
𝑃 𝐴 βˆͺ 𝐡 = 𝑃 𝐴 + 𝑃 𝐡 βˆ’ 𝑃(𝐴 ∩ 𝐡)
0.7 = 0.3 + 0.65 βˆ’ 𝑃(𝐴 ∩ 𝐡)
0.7 = 0.95 βˆ’ 𝑃(𝐴 ∩ 𝐡)
𝑃(𝐴 ∩ 𝐡) = 0.95 βˆ’ 0.70
𝑃(𝐴 ∩ 𝐡) = 0.25
MATH 110 Sec 13.2 Operations with Events Practice Exercises
The table below relates time consumers shop online/month by income.
What’s the probability that a randomly selected consumer either spends
0-2 hrs/month shopping online or has an annual income above $60,000?
(Round final Annual Income 10+ Hours 3-9 Hours 0-2 Hours Totals
answer to 2
Above $60000
182
171
121
474
decimal
$40000-60000
150
199
156
505
places.)
Below $40000
Totals
121
453
182
552
270
547
573
1552
MATH 110 Sec 13.2 Operations with Events Practice Exercises
The table below relates time consumers shop online/month by income.
What’s the probability that a randomly selected consumer either spends
0-2 hrs/month shopping online or has an annual income above $60,000?
(Round final Annual Income 10+ Hours 3-9 Hours 0-2 Hours Totals
answer to 2
Above $60000
182
171
121
474
decimal
$40000-60000
150
199
156
505
places.)
Below $40000
Totals
These are just subtotals.
121
453
182
552
270
547
573
1552
MATH 110 Sec 13.2 Operations with Events Practice Exercises
The table below relates time consumers shop online/month by income.
What’s the probability that a randomly selected consumer either spends
0-2 hrs/month shopping online or has an annual income above $60,000?
(Round final Annual Income 10+ Hours 3-9 Hours 0-2 Hours Totals
answer to 2
Above $60000
182
171
121
474
decimal
$40000-60000
150
199
156
505
places.)
Below $40000
Totals
These are just subtotals.
121
453
182
552
270
547
573
1552
This is the grand total
(of consumers surveyed)
MATH
13.2 Operations
with Events Practice Exercises
These
are110
theSec
numbers
of
consumers
who
shopped
The
table below
relates
time consumers shop online/month by income.
0-2the
hrs/month
online.
What’s
probability
that a randomly selected consumer either spends
0-2 hrs/month shopping online or has an annual income above $60,000?
(Round final Annual Income 10+ Hours 3-9 Hours 0-2 Hours Totals
answer to 2
Above $60000
182
171
121
474
decimal
$40000-60000
150
199
156
505
places.)
Below $40000
Totals
These are just subtotals.
121
453
182
552
270
547
573
1552
This is the grand total
(of consumers surveyed)
MATH
13.2 Operations
with
Events
Practice
Exercises
These
are the
numbers
of
These
are110
theSec
numbers
of
consumers
with annual
consumers
who
shopped
The
table below
relates
time consumers
shop online/month
by income.
$60,000.
0-2the
hrs/month
online.
What’s
probability
that a randomlyincomes
selectedabove
consumer
either spends
0-2 hrs/month shopping online or has an annual income above $60,000?
(Round final Annual Income 10+ Hours 3-9 Hours 0-2 Hours Totals
answer to 2
Above $60000
182
171
121
474
decimal
$40000-60000
150
199
156
505
places.)
Below $40000
Totals
These are just subtotals.
121
453
182
552
270
547
573
1552
This is the grand total
(of consumers surveyed)
MATH
110
Operations
with
Events
Practice
Because
it is
anSec
OR,13.2
we are
interested
in the
numbers
thatExercises
are inside of
either
or both
rectangles
The table below relates
time
consumers
shopbelow.
online/month by income.
What’s the probability that a randomly selected consumer either spends
0-2 hrs/month shopping online or has an annual income above $60,000?
(Round final Annual Income 10+ Hours 3-9 Hours 0-2 Hours Totals
answer to 2
Above $60000
182
171
121
474
decimal
$40000-60000
150
199
156
505
places.)
Below $40000
Totals
121
453
182
552
270
547
573
1552
This is the grand total
(of consumers surveyed)
MATH
110
Operations
with
Events
Practice
Because
it is
anSec
OR,13.2
we are
interested
in the
numbers
thatExercises
are inside of
either
or both
rectangles
The table below relates
time
consumers
shopbelow.
online/month by income.
What’s the probability that a randomly selected consumer either spends
0-2 hrs/month shopping online or has an annual income above $60,000?
(Round final Annual Income 10+ Hours 3-9 Hours 0-2 Hours Totals
answer to 2
Above $60000
182
171
121
474
decimal
$40000-60000
150
199
156
505
places.)
Below $40000
Totals
121
453
182
552
270
547
573
1552
182+171+121+156+270 = 900
This is the grand total
(of consumers surveyed)
MATH
110
Operations
with
Events
Practice
Because
it is
anSec
OR,13.2
we are
interested
in the
numbers
thatExercises
are inside of
either
or both
rectangles
The table below relates
time
consumers
shopbelow.
online/month by income.
What’s the probability that a randomly selected consumer either spends
0-2 hrs/month shopping online or has an annual income above $60,000?
(Round final Annual Income 10+ Hours 3-9 Hours 0-2 Hours Totals
answer to 2
Above $60000
182
171
121
474
decimal
$40000-60000
150
199
156
505
places.)
Below $40000
Totals
121
453
182
552
270
547
573
1552
182+171+121+156+270 = 900
This is the grand total
(of consumers surveyed)
MATH
110
Operations
with
Events
Practice
Because
it is
anSec
OR,13.2
we are
interested
in the
numbers
thatExercises
are inside of
either
or both
rectangles
The table below relates
time
consumers
shopbelow.
online/month by income.
What’s the probability that a randomly selected consumer either spends
0-2 hrs/month shopping online or has an annual income above $60,000?
(Round final Annual Income 10+ Hours 3-9 Hours 0-2 Hours Totals
answer to 2
Above $60000
182
171
121
474
decimal
$40000-60000
150
199
156
505
places.)
Below $40000
Totals
121
453
182
552
270
547
573
1552
182+171+121+156+270 = 900
0 βˆ’ 2 β„Žπ‘Ÿ π‘π‘’π‘Ÿ π‘šπ‘œ. 𝑂𝑅
900
𝑃
=
=0.58
1552
π‘–π‘›π‘π‘œπ‘šπ‘’ > $60000
This is the grand total
(of consumers surveyed)
MATH
110
Operations
with
Events
Practice
Because
it is
anSec
OR,13.2
we are
interested
in the
numbers
thatExercises
are inside of
either
or both
rectangles
The table below relates
time
consumers
shopbelow.
online/month by income.
What’s the probability that a randomly selected consumer either spends
0-2 hrs/month shopping online or has an annual income above $60,000?
(Round final Annual Income 10+ Hours 3-9 Hours 0-2 Hours Totals
answer to 2
Above $60000
182
171
121
474
decimal
$40000-60000
150
199
156
505
places.)
Below $40000
Totals
121
453
182
552
270
547
573
1552
182+171+121+156+270 = 900
0 βˆ’ 2 β„Žπ‘Ÿ π‘π‘’π‘Ÿ π‘šπ‘œ. 𝑂𝑅
900
𝑃
=
=0.58
1552
π‘–π‘›π‘π‘œπ‘šπ‘’ > $60000
This is the grand total
(of consumers surveyed)
MATH
110
Operations
with
Events
Practice
Because
it is
anSec
OR,13.2
we are
interested
in the
numbers
thatExercises
are inside of
either
or both
rectangles
The table below relates
time
consumers
shopbelow.
online/month by income.
What’s the probability that a randomly selected consumer either spends
0-2 hrs/month shopping online or has an annual income above $60,000?
(Round final Annual Income 10+ Hours 3-9 Hours 0-2 Hours Totals
answer to 2
Above $60000
182
171
121
474
decimal
$40000-60000
150
199
156
505
places.)
Below $40000
Totals
121
453
182
552
270
547
573
1552
182+171+121+156+270 = 900
0 βˆ’ 2 β„Žπ‘Ÿ π‘π‘’π‘Ÿ π‘šπ‘œ. 𝑂𝑅
900
This is the grand total
𝑃
=
= 0.58 (of
consumers surveyed)
1552
π‘–π‘›π‘π‘œπ‘šπ‘’ > $60000
MATH 110 Sec 13.2 Operations with Events Practice Exercises
The table below relates time consumers shop online/month by income.
What’s the probability that a randomly selected consumer either spends
0-2 hrs/month shopping online or has an annual income above $60,000?
(Round final Annual Income 10+ Hours 3-9 Hours 0-2 Hours Totals
answer to 2
Above $60000
182
171
121
474
decimal
$40000-60000
150
199
156
505
places.)
Below $40000
Totals
121
453
182
552
182+171+121+156+270 = 900
0 βˆ’ 2 β„Žπ‘Ÿ π‘π‘’π‘Ÿ π‘šπ‘œ. 𝑂𝑅
900
𝑃
=
= 0.58
1552
π‘–π‘›π‘π‘œπ‘šπ‘’ > $60000
270
547
573
1552
MATH 110 Sec 13.2 Operations with Events Practice Exercises
An employee at an electronics
store earns both a salary and a
monthly commission as a sales
rep. The table lists her
estimates of the probability of
earning various commissions
next month. What is the
probability that she will earn
no more than $1,999 in
commissions next month?
Commission
Less than $1000
$1000-$1249
$1250-$1499
$1500-$1749
$1750-$1999
$2000-$2249
$2250-$2499
$2500 or more
Probability
0.06
0.13
0.22
0.34
0.13
0.04
0.05
0.03
MATH 110 Sec 13.2 Operations with Events Practice Exercises
An employee at an electronics
store earns both a salary and a
monthly commission as a sales
rep. The table lists her
estimates of the probability of
earning various commissions
next month. What is the
probability that she will earn
no more than $1,999 in
commissions next month?
Commission
Less than $1000
$1000-$1249
$1250-$1499
$1500-$1749
$1750-$1999
$2000-$2249
$2250-$2499
$2500 or more
Probability
0.06
0.13
0.22
0.34
0.13
0.04
0.05
0.03
MATH 110 Sec 13.2 Operations with Events Practice Exercises
An employee at an electronics
store earns both a salary and a
monthly commission as a sales
rep. The table lists her
estimates of the probability of
earning various commissions
next month. What is the
probability that she will earn
no more than $1,999 in
commissions next month?
𝑃
Commission
Less than $1000
$1000-$1249
$1250-$1499
$1500-$1749
$1750-$1999
$2000-$2249
$2250-$2499
$2500 or more
Probability
0.06
0.13
0.22
0.34
0.13
0.04
0.05
0.03
π‘’π‘Žπ‘Ÿπ‘›π‘  π‘›π‘œ π‘šπ‘œπ‘Ÿπ‘’ π‘‘β„Žπ‘Žπ‘›
= 0.06 + 0.13 + 0.22 + 0.34 + 0.13 =0.58
$1999 𝑛𝑒π‘₯𝑑 π‘šπ‘œπ‘›π‘‘β„Ž
MATH 110 Sec 13.2 Operations with Events Practice Exercises
An employee at an electronics
store earns both a salary and a
monthly commission as a sales
rep. The table lists her
estimates of the probability of
earning various commissions
next month. What is the
probability that she will earn
no more than $1,999 in
commissions next month?
𝑃
Commission
Less than $1000
$1000-$1249
$1250-$1499
$1500-$1749
$1750-$1999
$2000-$2249
$2250-$2499
$2500 or more
Probability
0.06
0.13
0.22
0.34
0.13
0.04
0.05
0.03
π‘’π‘Žπ‘Ÿπ‘›π‘  π‘›π‘œ π‘šπ‘œπ‘Ÿπ‘’ π‘‘β„Žπ‘Žπ‘›
= 0.06 + 0.13 + 0.22 + 0.34 + 0.13 =0.58
$1999 𝑛𝑒π‘₯𝑑 π‘šπ‘œπ‘›π‘‘β„Ž
MATH 110 Sec 13.2 Operations with Events Practice Exercises
An employee at an electronics
store earns both a salary and a
monthly commission as a sales
rep. The table lists her
estimates of the probability of
earning various commissions
next month. What is the
probability that she will earn
no more than $1,999 in
commissions next month?
𝑃
Commission
Less than $1000
$1000-$1249
$1250-$1499
$1500-$1749
$1750-$1999
$2000-$2249
$2250-$2499
$2500 or more
Probability
0.06
0.13
0.22
0.34
0.13
0.04
0.05
0.03
π‘’π‘Žπ‘Ÿπ‘›π‘  π‘›π‘œ π‘šπ‘œπ‘Ÿπ‘’ π‘‘β„Žπ‘Žπ‘›
= 0.06 + 0.13 + 0.22 + 0.34 + 0.13 =0.58
$1999 𝑛𝑒π‘₯𝑑 π‘šπ‘œπ‘›π‘‘β„Ž
MATH 110 Sec 13.2 Operations with Events Practice Exercises
An employee at an electronics
store earns both a salary and a
monthly commission as a sales
rep. The table lists her
estimates of the probability of
earning various commissions
next month. What is the
probability that she will earn
no more than $1,999 in
commissions next month?
𝑃
Commission
Less than $1000
$1000-$1249
$1250-$1499
$1500-$1749
$1750-$1999
$2000-$2249
$2250-$2499
$2500 or more
Probability
0.06
0.13
0.22
0.34
0.13
0.04
0.05
0.03
π‘’π‘Žπ‘Ÿπ‘›π‘  π‘›π‘œ π‘šπ‘œπ‘Ÿπ‘’ π‘‘β„Žπ‘Žπ‘›
= 0.06 + 0.13 + 0.22 + 0.34 + 0.13 = 0.58
$1999 𝑛𝑒π‘₯𝑑 π‘šπ‘œπ‘›π‘‘β„Ž
MATH 110 Sec 13.2 Operations with Events Practice Exercises
The windshield wipers on a car have not been working properly. The
probability that the car needs a new motor is 0.55, that it needs a new
switch is 0.45 and the probability that it both is 0.15. What is the
probability that the car needs neither a new motor nor a new switch?
MATH 110 Sec 13.2 Operations with Events Practice Exercises
The windshield wipers on a car have not been working properly. The
probability that the car needs a new motor is 0.55, that it needs a new
switch is 0.45 and the probability that it both is 0.15. What is the
probability that the car needs neither a new motor nor a new switch?
Venn diagrams can be useful in solving problems like this.
MATH 110 Sec 13.2 Operations with Events Practice Exercises
The windshield wipers on a car have not been working properly. The
probability that the car needs a new motor is 0.55, that it needs a new
switch is 0.45 and the probability that it both is 0.15. What is the
probability that the car needs neither a new motor nor a new switch?
Venn diagrams can be useful in solving problems like this.
M
S
M is β€˜needs new motor’.
S is β€˜needs new switch’.
MATH 110 Sec 13.2 Operations with Events Practice Exercises
The windshield wipers on a car have not been working properly. The
probability that the car needs a new motor is 0.55, that it needs a new
switch is 0.45 and the probability that it both is 0.15. What is the
probability that the car needs neither a new motor nor a new switch?
Venn diagrams can be useful in solving problems like this.
M
S
0.15
M is β€˜needs new motor’.
S is β€˜needs new switch’.
MATH 110 Sec 13.2 Operations with Events Practice Exercises
The windshield wipers on a car have not been working properly. The
probability that the car needs a new motor is 0.55, that it needs a new
switch is 0.45 and the probability that it both is 0.15. What is the
probability that the car needs neither a new motor nor a new switch?
Venn diagrams can be useful in solving problems like this.
M
S
0.15
M is β€˜needs new motor’.
S is β€˜needs new switch’.
MATH 110 Sec 13.2 Operations with Events Practice Exercises
The windshield wipers on a car have not been working properly. The
probability that the car needs a new motor is 0.55, that it needs a new
switch is 0.45 and the probability that it both is 0.15. What is the
probability that the car needs neither a new motor nor a new switch?
Venn diagrams can be useful in solving problems like this.
M
S
0.15
M is β€˜needs new motor’.
S is β€˜needs new switch’.
MATH 110 Sec 13.2 Operations with Events Practice Exercises
The windshield wipers on a car have not been working properly. The
probability that the car needs a new motor is 0.55, that it needs a new
switch is 0.45 and the probability that it both is 0.15. What is the
probability that the car needs neither a new motor nor a new switch?
Venn diagrams can be useful in solving problems like this.
M
S
0.15
M is β€˜needs new motor’.
S is β€˜needs new switch’.
MATH 110 Sec 13.2 Operations with Events Practice Exercises
The windshield wipers on a car have not been working properly. The
probability that the car needs a new motor is 0.55, that it needs a new
switch is 0.45 and the probability that it both is 0.15. What is the
probability that the car needs neither a new motor nor a new switch?
Venn diagrams can be useful in solving problems like this.
M
S
0.15
M is β€˜needs new motor’.
S is β€˜needs new switch’.
MATH 110 Sec 13.2 Operations with Events Practice Exercises
The windshield wipers on a car have not been working properly. The
probability that the car needs a new motor is 0.55, that it needs a new
switch is 0.45 and the probability that it both is 0.15. What is the
probability that the car needs neither a new motor nor a new switch?
Venn diagrams can be useful in solving problems like this.
M
S
0.15
M is β€˜needs new motor’.
S is β€˜needs new switch’.
MATH 110 Sec 13.2 Operations with Events Practice Exercises
The windshield wipers on a car have not been working properly. The
probability that the car needs a new motor is 0.55, that it needs a new
switch is 0.45 and the probability that it both is 0.15. What is the
probability that the car needs neither a new motor nor a new switch?
Venn diagrams can be useful in solving problems like this.
M
S
0.15
M is β€˜needs new motor’.
S is β€˜needs new switch’.
MATH 110 Sec 13.2 Operations with Events Practice Exercises
The windshield wipers on a car have not been working properly. The
probability that the car needs a new motor is 0.55, that it needs a new
switch is 0.45 and the probability that it both is 0.15. What is the
probability that the car needs neither a new motor nor a new switch?
Venn diagrams can be useful in solving problems like this.
M
S
0.15
M is β€˜needs new motor’.
S is β€˜needs new switch’.
MATH 110 Sec 13.2 Operations with Events Practice Exercises
The windshield wipers on a car have not been working properly. The
probability that the car needs a new motor is 0.55, that it needs a new
switch is 0.45 and the probability that it both is 0.15. What is the
probability that the car needs neither a new motor nor a new switch?
Venn diagrams can be useful in solving problems like this.
M
M is β€˜needs new motor’.
S is β€˜needs new switch’.
S
0.15
=1
?
MATH 110 Sec 13.2 Operations with Events Practice Exercises
The windshield wipers on a car have not been working properly. The
probability that the car needs a new motor is 0.55, that it needs a new
switch is 0.45 and the probability that it both is 0.15. What is the
probability that the car needs neither a new motor nor a new switch?
Venn diagrams can be useful in solving problems like this.
M
M is β€˜needs new motor’.
S is β€˜needs new switch’.
S
0.15
=1
0.15
MATH 110 Sec 13.2 Operations with Events Practice Exercises
The windshield wipers on a car have not been working properly. The
probability that the car needs a new motor is 0.55, that it needs a new
switch is 0.45 and the probability that it both is 0.15. What is the
probability that the car needs neither a new motor nor a new switch?
Venn diagrams can be useful in solving problems like this.
M
M is β€˜needs new motor’.
S is β€˜needs new switch’.
S
0.15
0.15
MATH 110 Sec 13.2 Operations with Events Practice Exercises
The windshield wipers on a car have not been working properly. The
probability that the car needs a new motor is 0.55, that it needs a new
switch is 0.45 and the probability that it both is 0.15. What is the
probability that the car needs neither a new motor nor a new switch?
Venn diagrams can be useful in solving problems like this.
M
M is β€˜needs new motor’.
S is β€˜needs new switch’.
S
𝑃
0.15
0.15
𝑛𝑒𝑒𝑑𝑠
=
π‘›π‘’π‘–π‘‘β„Žπ‘’π‘Ÿ
MATH 110 Sec 13.2 Operations with Events Practice Exercises
The windshield wipers on a car have not been working properly. The
probability that the car needs a new motor is 0.55, that it needs a new
switch is 0.45 and the probability that it both is 0.15. What is the
probability that the car needs neither a new motor nor a new switch?
Venn diagrams can be useful in solving problems like this.
M
M is β€˜needs new motor’.
S is β€˜needs new switch’.
S
𝑃
0.15
0.15
𝑛𝑒𝑒𝑑𝑠
= 0.15
π‘›π‘’π‘–π‘‘β„Žπ‘’π‘Ÿ
MATH 110 Sec 13.2 Operations with Events Practice Exercises
The windshield wipers on a car have not been working properly. The
probability that the car needs a new motor is 0.55, that it needs a new
switch is 0.45 and the probability that it both is 0.15. What is the
probability that the car needs neither a new motor nor a new switch?
Venn diagrams can be useful in solving problems like this.
M
M is β€˜needs new motor’.
S is β€˜needs new switch’.
S
𝑃
0.15
0.15
𝑛𝑒𝑒𝑑𝑠
= 0.15
π‘›π‘’π‘–π‘‘β„Žπ‘’π‘Ÿ
MATH 110 Sec 13.2 Operations with Events Practice Exercises
A study of 205 randomly selected students examined the relationship
between satisfaction with academic advising and academic success. Of
the 62 students on academic probation, 22 are not satisfied with their
advising. But, only 25 of the students not on academic probation are
dissatisfied with their advising. What’s the probability that a randomly
selected student is on academic probation and is satisfied with advising?
Round final answer to 2 decimal places.
MATH 110 Sec 13.2 Operations with Events Practice Exercises
A study of 205 randomly selected students examined the relationship
between satisfaction with academic advising and academic success. Of
the 62 students on academic probation, 22 are not satisfied with their
advising. But, only 25 of the students not on academic probation are
dissatisfied with their advising. What’s the probability that a randomly
selected student is on academic probation and is satisfied with advising?
Round final answer to 2 decimal places.
S
P
S
P
S is β€˜satisfied with advising’.
P is β€˜is on academic probation’.
MATH 110 Sec 13.2 Operations with Events Practice Exercises
In other
words,
there students
are 62 students
in circle
P
A study of 205
randomly
selected
examined
the relationship
but
22 academic
of those are
NOT inand
circle
S.
between satisfaction
with
advising
academic
success. Of
the 62 students on academic probation, 22 are not satisfied with their
advising. But, only 25 of the students not on academic probation are
dissatisfied with their advising. What’s the probability that a randomly
selected student is on academic probation and is satisfied with advising?
Round final answer to 2 decimal places.
S
P
S
P
S is β€˜satisfied with advising’.
P is β€˜is on academic probation’.
MATH 110 Sec 13.2 Operations with Events Practice Exercises
In other
words,
there students
are 62 students
in circle
P
A study of 205
randomly
selected
examined
the relationship
but
22 academic
of those are
NOT inand
circle
S.
between satisfaction
with
advising
academic
success. Of
the 62 students on academic probation, 22 are not satisfied with their
advising. But, only 25 of the students not on academic probation are
dissatisfied with their advising. What’s the probability that a randomly
selected student is on academic probation and is satisfied with advising?
Round final answer to 2 decimal places.
S
P
S
P
S is β€˜satisfied with advising’.
P is β€˜is on academic probation’.
22
MATH 110 Sec 13.2 Operations with Events Practice Exercises
In other
words,
there students
are 62 students
in circle
P
A study of 205
randomly
selected
examined
the relationship
but
22 academic
of those are
NOT inand
circle
S.
between satisfaction
with
advising
academic
success. Of
the 62 students on academic probation, 22 are not satisfied with their
advising. But, only 25 of the students not on academic probation are
dissatisfied with their advising. What’s the probability that a randomly
selected student is on academic probation and is satisfied with advising?
Round final answer to 2 decimal places.
S
P
S
P
S is β€˜satisfied with advising’.
P is β€˜is on academic probation’.
22
MATH 110 Sec 13.2 Operations with Events Practice Exercises
In other
words,
there students
are 62 students
in circle
P
A study of 205
randomly
selected
examined
the relationship
but
22 academic
of those are
NOT inand
circle
S.
between satisfaction
with
advising
academic
success. Of
the 62 students on academic probation, 22 are not satisfied with their
advising. But, only 25 of the students not on academic probation are
dissatisfied with their advising. What’s the probability that a randomly
selected student is on academic probation and is satisfied with advising?
Round final answer to 2 decimal places.
S
P
S
P
S is β€˜satisfied with advising’.
P is β€˜is on academic probation’.
22
But if there are 62 students
in circle P, then there must
be 40 students here.
MATH 110 Sec 13.2 Operations with Events Practice Exercises
In other
words,
there students
are 62 students
in circle
P
A study of 205
randomly
selected
examined
the relationship
but
22 academic
of those are
NOT inand
circle
S.
between satisfaction
with
advising
academic
success. Of
the 62 students on academic probation, 22 are not satisfied with their
advising. But, only 25 of the students not on academic probation are
dissatisfied with their advising. What’s the probability that a randomly
selected student is on academic probation and is satisfied with advising?
Round final answer to 2 decimal places.
S
P
S
P
S is β€˜satisfied with advising’.
40
P is β€˜is on academic probation’.
22
But if there are 62 students
in circle P, then there must
be 40 students here.
MATH 110 Sec 13.2 Operations with Events Practice Exercises
A study of 205 randomly selected students examined the relationship
between satisfaction with academic advising and academic success. Of
the 62 students on academic probation, 22 are not satisfied with their
advising. But, only 25 of the students not on academic probation are
dissatisfied with their advising. What’s the probability that a randomly
selected student is on academic probation and is satisfied with advising?
Round final answer to 2 decimal places.
S
P
S
P
S is β€˜satisfied with advising’.
40
P is β€˜is on academic probation’.
22
MATH 110 Sec 13.2 Operations with Events Practice Exercises
So, only 25
students
not in examined the relationship
A study of 205 randomly
selected
students
circle
P are
also notadvising
in S. and academic success. Of
between satisfaction
with
academic
the 62 students on academic probation, 22 are not satisfied with their
advising. But, only 25 of the students not on academic probation are
dissatisfied with their advising. What’s the probability that a randomly
selected student is on academic probation and is satisfied with advising?
Round final answer to 2 decimal places.
S
P
S
P
S is β€˜satisfied with advising’.
40
P is β€˜is on academic probation’.
22
MATH 110 Sec 13.2 Operations with Events Practice Exercises
So, only 25
students
not in examined the relationship
A study of 205 randomly
selected
students
circle
P are
also notadvising
in S. and academic success. Of
between satisfaction
with
academic
the 62 students on academic probation, 22 are not satisfied with their
advising. But, only 25 of the students not on academic probation are
dissatisfied with their advising. What’s the probability that a randomly
selected student is on academic probation and is satisfied with advising?
Round final answer to 2 decimal places.
S
P
S
P
S is β€˜satisfied with advising’.
40
P is β€˜is on academic probation’.
22
25
MATH 110 Sec 13.2 Operations with Events Practice Exercises
A study of 205 randomly selected students examined the relationship
between satisfaction with academic advising and academic success. Of
the 62 students on academic probation, 22 are not satisfied with their
advising. But, only 25 of the students not on academic probation are
dissatisfied with their advising. What’s the probability that a randomly
selected student is on academic probation and is satisfied with advising?
Round final answer to 2 decimal places.
S
P
S
P
40
22
There are 205 students in all and we
have already accounted for
40+22+25 = 87 of them.
25 So, there must be 205-87 = 118 here.
MATH 110 Sec 13.2 Operations with Events Practice Exercises
A study of 205 randomly selected students examined the relationship
between satisfaction with academic advising and academic success. Of
the 62 students on academic probation, 22 are not satisfied with their
advising. But, only 25 of the students not on academic probation are
dissatisfied with their advising. What’s the probability that a randomly
selected student is on academic probation and is satisfied with advising?
Round final answer to 2 decimal places.
S
P
S
P
40
22
There are 205 students in all and we
have already accounted for
40+22+25 = 87 of them.
25 So, there must be 205-87 = 118 here.
MATH 110 Sec 13.2 Operations with Events Practice Exercises
A study of 205 randomly selected students examined the relationship
between satisfaction with academic advising and academic success. Of
the 62 students on academic probation, 22 are not satisfied with their
advising. But, only 25 of the students not on academic probation are
dissatisfied with their advising. What’s the probability that a randomly
selected student is on academic probation and is satisfied with advising?
Round final answer to 2 decimal places.
S
P
S
P
118
40
22
There are 205 students in all and we
have already accounted for
40+22+25 = 87 of them.
25 So, there must be 205-87 = 118 here.
MATH 110 Sec 13.2 Operations with Events Practice Exercises
A study of 205 randomly selected students examined the relationship
between satisfaction with academic advising and academic success. Of
the 62 students on academic probation, 22 are not satisfied with their
advising. But, only 25 of the students not on academic probation are
dissatisfied with their advising. What’s the probability that a randomly
selected student is on academic probation and is satisfied with advising?
Round final answer to 2 decimal places.
S
P
S
P
S is β€˜satisfied with advising’.
40
118
P is β€˜is on academic probation’.
22
25
MATH 110 Sec 13.2 Operations with Events Practice Exercises
A study of 205 randomly selected students examined the relationship
between satisfaction with academic advising and academic success. Of
the 62 students on academic probation, 22 are not satisfied with their
advising. But, only 25 of the students not on academic probation are
dissatisfied with their advising. What’s the probability that a randomly
selected student is on academic probation and is satisfied with advising?
Round final answer to 2 decimal places.
S
P
S
P
S is β€˜satisfied with advising’.
40
118
P is β€˜is on academic probation’.
22
25
MATH 110 Sec 13.2 Operations with Events Practice Exercises
A study of 205 randomly selected students examined the relationship
between satisfaction with academic advising and academic success. Of
the 62 students on academic probation, 22 are not satisfied with their
advising. But, only 25 of the students not on academic probation are
dissatisfied with their advising. What’s the probability that a randomly
selected student is on academic probation and is satisfied with advising?
Round final answer to 2 decimal places.
S
P
S
P
S is β€˜satisfied with advising’.
40
118
P is β€˜is on academic probation’.
22
25
MATH 110 Sec 13.2 Operations with Events Practice Exercises
A study of 205 randomly selected students examined the relationship
between satisfaction with academic advising and academic success. Of
the 62 students on academic probation, 22 are not satisfied with their
advising. But, only 25 of the students not on academic probation are
dissatisfied with their advising. What’s the probability that a randomly
selected student is on academic probation and is satisfied with advising?
Round final answer to 2 decimal places.
S
P
S
P
S is β€˜satisfied with advising’.
40
118
P is β€˜is on academic probation’.
22
40
π‘œπ‘› π‘π‘Ÿπ‘œπ‘π‘Žπ‘‘π‘–π‘œπ‘›
= 0.20
25 𝑃 π‘Žπ‘›π‘‘ π‘ π‘Žπ‘‘π‘–π‘ π‘“π‘–π‘’π‘‘ =
205
MATH 110 Sec 13.2 Operations with Events Practice Exercises
A company shipped 57 cameras to a store. Two of these are defective.
If the store sold 20 cameras before discovering that some were defective,
what is the probability that at least one defective camera was sold?
MATH 110 Sec 13.2 Operations with Events Practice Exercises
A company shipped 57 cameras to a store. Two of these are defective.
If the store sold 20 cameras before discovering that some were defective,
what is the probability that at least one defective camera was sold?
HINT:
π‘ͺ(πŸ“πŸ“,𝟐𝟎)
π‘ͺ(πŸ“πŸ•,𝟐𝟎)
is the probability that no defective cameras were sold.
MATH 110 Sec 13.2 Operations with Events Practice Exercises
A company shipped 57 cameras to a store. Two of these are defective.
If the store sold 20 cameras before discovering that some were defective,
what is the probability that at least one defective camera was sold?
HINT:
π‘ͺ(πŸ“πŸ“,𝟐𝟎)
π‘ͺ(πŸ“πŸ•,𝟐𝟎)
is the probability that no defective cameras were sold.
COMPLEMENT RULE: 𝑃 𝐸 = 1 βˆ’ 𝑃(𝐸 β€² )
MATH 110 Sec 13.2 Operations with Events Practice Exercises
A company shipped 57 cameras to a store. Two of these are defective.
If the store sold 20 cameras before discovering that some were defective,
what is the probability that at least one defective camera was sold?
HINT:
π‘ͺ(πŸ“πŸ“,𝟐𝟎)
π‘ͺ(πŸ“πŸ•,𝟐𝟎)
is the probability that no defective cameras were sold.
COMPLEMENT RULE: 𝑃 𝐸 = 1 βˆ’ 𝑃(𝐸 β€² )
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ 1
π‘›π‘œ 𝑑𝑒𝑓𝑒𝑐𝑑𝑖𝑣𝑒
𝑃
=1βˆ’π‘ƒ
𝑑𝑒𝑓𝑒𝑐𝑑𝑖𝑣𝑒 π‘ π‘œπ‘™π‘‘
π‘ π‘œπ‘™π‘‘
MATH 110 SecBut
13.2the
Operations
with
Events
Practice
hint told us
how
to calculate
thisExercises
probability.
A company shipped 57 cameras to a store. Two of these are defective.
If the store sold 20 cameras before discovering that some were defective,
what is the probability that at least one defective camera was sold?
HINT:
π‘ͺ(πŸ“πŸ“,𝟐𝟎)
π‘ͺ(πŸ“πŸ•,𝟐𝟎)
is the probability that no defective cameras were sold.
COMPLEMENT RULE: 𝑃 𝐸 = 1 βˆ’ 𝑃(𝐸 β€² )
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ 1
π‘›π‘œ 𝑑𝑒𝑓𝑒𝑐𝑑𝑖𝑣𝑒
𝑃
=1βˆ’π‘ƒ
𝑑𝑒𝑓𝑒𝑐𝑑𝑖𝑣𝑒 π‘ π‘œπ‘™π‘‘
π‘ π‘œπ‘™π‘‘
MATH 110 SecBut
13.2the
Operations
with
Events
Practice
hint told us
how
to calculate
thisExercises
probability.
A company shipped 57 cameras to a store. Two of these are defective.
If the store sold 20 cameras before discovering that some were defective,
what is the probability that at least one defective camera was sold?
HINT:
π‘ͺ(πŸ“πŸ“,𝟐𝟎)
π‘ͺ(πŸ“πŸ•,𝟐𝟎)
is the probability that no defective cameras were sold.
COMPLEMENT RULE: 𝑃 𝐸 = 1 βˆ’ 𝑃(𝐸 β€² )
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ 1
π‘›π‘œ 𝑑𝑒𝑓𝑒𝑐𝑑𝑖𝑣𝑒
𝑃
=1βˆ’π‘ƒ
𝑑𝑒𝑓𝑒𝑐𝑑𝑖𝑣𝑒 π‘ π‘œπ‘™π‘‘
π‘ π‘œπ‘™π‘‘
MATH 110 SecBut
13.2the
Operations
with
Events
Practice
hint told us
how
to calculate
thisExercises
probability.
𝐢(55,20)
A company shipped 57 cameras
to a store. Two
of these are defective.
π‘›π‘œ 𝑑𝑒𝑓𝑒𝑐𝑑𝑖𝑣𝑒
𝑃 before discovering
= that some were
= 0.417
If the store sold 20 cameras
defective,
𝐢(57,20)
π‘ π‘œπ‘™π‘‘
what is the probability that at least one defective camera was sold?
HINT:
π‘ͺ(πŸ“πŸ“,𝟐𝟎)
π‘ͺ(πŸ“πŸ•,𝟐𝟎)
is the probability that no defective cameras were sold.
COMPLEMENT RULE: 𝑃 𝐸 = 1 βˆ’ 𝑃(𝐸 β€² )
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ 1
π‘›π‘œ 𝑑𝑒𝑓𝑒𝑐𝑑𝑖𝑣𝑒
𝑃
=1βˆ’π‘ƒ
𝑑𝑒𝑓𝑒𝑐𝑑𝑖𝑣𝑒 π‘ π‘œπ‘™π‘‘
π‘ π‘œπ‘™π‘‘
MATH 110 SecBut
13.2the
Operations
with
Events
Practice
hint told us
how
to calculate
thisExercises
probability.
𝐢(55,20)
A company shipped 57 cameras
to a store. Two
of these are defective.
π‘›π‘œ 𝑑𝑒𝑓𝑒𝑐𝑑𝑖𝑣𝑒
𝑃 before discovering
= that some were
= 0.417
If the store sold 20 cameras
defective,
𝐢(57,20)
π‘ π‘œπ‘™π‘‘
what is the probability that at least one defective camera was sold?
HINT:
π‘ͺ(πŸ“πŸ“,𝟐𝟎)
π‘ͺ(πŸ“πŸ•,𝟐𝟎)
is the probability that no defective cameras were sold.
COMPLEMENT RULE: 𝑃 𝐸 = 1 βˆ’ 𝑃(𝐸 β€² )
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ 1
π‘›π‘œ 𝑑𝑒𝑓𝑒𝑐𝑑𝑖𝑣𝑒
𝑃
=1βˆ’π‘ƒ
𝑑𝑒𝑓𝑒𝑐𝑑𝑖𝑣𝑒 π‘ π‘œπ‘™π‘‘
π‘ π‘œπ‘™π‘‘
MATH 110 SecBut
13.2the
Operations
with
Events
Practice
hint told us
how
to calculate
thisExercises
probability.
𝐢(55,20)
A company shipped 57 cameras
to a store. Two
of these are defective.
π‘›π‘œ 𝑑𝑒𝑓𝑒𝑐𝑑𝑖𝑣𝑒
𝑃 before discovering
= that some were
= 0.417
If the store sold 20 cameras
defective,
𝐢(57,20)
π‘ π‘œπ‘™π‘‘
what is the probability that at least one defective camera was sold?
HINT:
π‘ͺ(πŸ“πŸ“,𝟐𝟎)
π‘ͺ(πŸ“πŸ•,𝟐𝟎)
is the probability that no defective cameras were sold.
COMPLEMENT RULE: 𝑃 𝐸 = 1 βˆ’ 𝑃(𝐸 β€² )
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ 1
π‘›π‘œ 𝑑𝑒𝑓𝑒𝑐𝑑𝑖𝑣𝑒
𝑃
=1βˆ’π‘ƒ
𝑑𝑒𝑓𝑒𝑐𝑑𝑖𝑣𝑒 π‘ π‘œπ‘™π‘‘
π‘ π‘œπ‘™π‘‘
Keystrokes: 5
557
205557
20
MATH 110 SecBut
13.2the
Operations
with
Events
Practice
hint told us
how
to calculate
thisExercises
probability.
𝐢(55,20)
A company shipped 57 cameras
to a store. Two
of these are defective.
π‘›π‘œ 𝑑𝑒𝑓𝑒𝑐𝑑𝑖𝑣𝑒
𝑃 before discovering
= that some were
= 0.417
If the store sold 20 cameras
defective,
𝐢(57,20)
π‘ π‘œπ‘™π‘‘
what is the probability that at least one defective camera was sold?
HINT:
π‘ͺ(πŸ“πŸ“,𝟐𝟎)
π‘ͺ(πŸ“πŸ•,𝟐𝟎)
is the probability that no defective cameras were sold.
COMPLEMENT RULE: 𝑃 𝐸 = 1 βˆ’ 𝑃(𝐸 β€² )
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ 1
π‘›π‘œ 𝑑𝑒𝑓𝑒𝑐𝑑𝑖𝑣𝑒
𝑃
=1βˆ’π‘ƒ
𝑑𝑒𝑓𝑒𝑐𝑑𝑖𝑣𝑒 π‘ π‘œπ‘™π‘‘
π‘ π‘œπ‘™π‘‘
Keystrokes: 5
557
205557
20
MATH 110 SecBut
13.2the
Operations
with
Events
Practice
hint told us
how
to calculate
thisExercises
probability.
𝐢(55,20)
A company shipped 57 cameras
to a store. Two
of these are defective.
π‘›π‘œ 𝑑𝑒𝑓𝑒𝑐𝑑𝑖𝑣𝑒
𝑃 before discovering
= that some were
= 0.417
If the store sold 20 cameras
defective,
𝐢(57,20)
π‘ π‘œπ‘™π‘‘
what is the probability that at least one defective camera was sold?
HINT:
π‘ͺ(πŸ“πŸ“,𝟐𝟎)
π‘ͺ(πŸ“πŸ•,𝟐𝟎)
is the probability that no defective cameras were sold.
COMPLEMENT RULE: 𝑃 𝐸 = 1 βˆ’ 𝑃(𝐸 β€² )
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ 1
π‘›π‘œ 𝑑𝑒𝑓𝑒𝑐𝑑𝑖𝑣𝑒
𝑃
=1βˆ’π‘ƒ
𝑑𝑒𝑓𝑒𝑐𝑑𝑖𝑣𝑒 π‘ π‘œπ‘™π‘‘
π‘ π‘œπ‘™π‘‘
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ 1
𝑃
= 1 βˆ’0.417
𝑑𝑒𝑓𝑒𝑐𝑑𝑖𝑣𝑒 π‘ π‘œπ‘™π‘‘
Keystrokes: 5
557
205557
20
MATH 110 SecBut
13.2the
Operations
with
Events
Practice
hint told us
how
to calculate
thisExercises
probability.
𝐢(55,20)
A company shipped 57 cameras
to a store. Two
of these are defective.
π‘›π‘œ 𝑑𝑒𝑓𝑒𝑐𝑑𝑖𝑣𝑒
𝑃 before discovering
= that some were
= 0.417
If the store sold 20 cameras
defective,
𝐢(57,20)
π‘ π‘œπ‘™π‘‘
what is the probability that at least one defective camera was sold?
HINT:
π‘ͺ(πŸ“πŸ“,𝟐𝟎)
π‘ͺ(πŸ“πŸ•,𝟐𝟎)
is the probability that no defective cameras were sold.
COMPLEMENT RULE: 𝑃 𝐸 = 1 βˆ’ 𝑃(𝐸 β€² )
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ 1
π‘›π‘œ 𝑑𝑒𝑓𝑒𝑐𝑑𝑖𝑣𝑒
𝑃
=1βˆ’π‘ƒ
𝑑𝑒𝑓𝑒𝑐𝑑𝑖𝑣𝑒 π‘ π‘œπ‘™π‘‘
π‘ π‘œπ‘™π‘‘
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ 1
𝑃
= 1 βˆ’0.417
𝑑𝑒𝑓𝑒𝑐𝑑𝑖𝑣𝑒 π‘ π‘œπ‘™π‘‘
= 0.583
MATH 110 Sec 13.2 Operations with Events Practice Exercises
A sushi shop prepared 23 portions of seafood, 2 of which stayed out too
long and spoiled. If 18 of the 23 portions are served randomly, what is
the probability that at least one customer will receive spoiled food?
Give final answer as a simplified fraction.
MATH 110 Sec 13.2 Operations with Events Practice Exercises
A sushi shop prepared 23 portions of seafood, 2 of which stayed out too
long and spoiled. If 18 of the 23 portions are served randomly, what is
the probability that at least one customer will receive spoiled food?
Give final answer as a simplified fraction.
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘
As in the previous problem, it is easier to find 𝑃
first
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
MATH 110 Sec 13.2 Operations with Events Practice Exercises
A sushi shop prepared 23 portions of seafood, 2 of which stayed out too
long and spoiled. If 18 of the 23 portions are served randomly, what is
the probability that at least one customer will receive spoiled food?
Give final answer as a simplified fraction.
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘
As in the previous problem, it is easier to find 𝑃
first
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
and then use the Complement Rule.
MATH 110 Sec 13.2 Operations with Events Practice Exercises
A sushi shop prepared 23 portions of seafood, 2 of which stayed out too
long and spoiled. If 18 of the 23 portions are served randomly, what is
the probability that at least one customer will receive spoiled food?
Give final answer as a simplified fraction.
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘
As in the previous problem, it is easier to find 𝑃
first
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
and then use the Complement Rule.
COMPLEMENT RULE: 𝑃 𝐸 = 1 βˆ’ 𝑃(𝐸 β€² )
MATH 110 Sec 13.2 Operations with Events Practice Exercises
A sushi shop prepared 23 portions of seafood, 2 of which stayed out too
long and spoiled. If 18 of the 23 portions are served randomly, what is
the probability that at least one customer will receive spoiled food?
Give final answer as a simplified fraction.
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘
As in the previous problem, it is easier to find 𝑃
first
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
and then use the Complement Rule.
COMPLEMENT RULE: 𝑃 𝐸 = 1 βˆ’ 𝑃(𝐸 β€² )
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ π‘œπ‘›π‘’ π‘”π‘œπ‘‘
𝑃
=1βˆ’π‘ƒ
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
MATH 110 Sec 13.2 Operations with Events Practice Exercises
A sushi shop prepared 23 portions of seafood, 2 of which stayed out too
long and spoiled. If 18 of the 23 portions are served randomly, what is
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘
the probability that at least one customer
will receive spoiled food?
let’s
find 𝑃as a simplified fraction.
.
GiveSo,
final
answer
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘
As in the previous problem, it is easier to find 𝑃
first
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
and then use the Complement Rule.
COMPLEMENT RULE: 𝑃 𝐸 = 1 βˆ’ 𝑃(𝐸 β€² )
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ π‘œπ‘›π‘’ π‘”π‘œπ‘‘
𝑃
=1βˆ’π‘ƒ
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
MATH 110 Sec 13.2 Operations with Events Practice Exercises
A sushi shop prepared 23 portions of seafood, 2 of which stayed out too
long and spoiled. If 18 of the 23 portions are served randomly, what is
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘
the probability that at least one customer
will receive spoiled food?
let’s
find 𝑃as a simplified fraction.
.
GiveSo,
final
answer
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘to be
(The
hint
on
the
previous
problem
tells
us
exactly
what
needs
As in the previous problem, it is easier to find 𝑃
first
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘
done here, but we will ignore that hint and just work it all out.)
and then use the Complement Rule.
COMPLEMENT RULE: 𝑃 𝐸 = 1 βˆ’ 𝑃(𝐸 β€² )
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ π‘œπ‘›π‘’ π‘”π‘œπ‘‘
𝑃
=1βˆ’π‘ƒ
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
MATH 110 Sec 13.2 Operations with Events Practice Exercises
A sushi shop prepared 23 portions of seafood, 2 of which stayed out too
long and spoiled. If 18 of the 23 portions are served randomly, what is
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘
the probability that at least one customer
will receive spoiled food?
let’s
find 𝑃as a simplified fraction.
.
GiveSo,
final
answer
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘to be
(The
hint
on
the
previous
problem
tells
us
exactly
what
needs
As in the previous problem, it is easier to find 𝑃
first
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘
done here, but we will ignore that hint and just work it all out.)
and
thenthe
useBasic
the Complement
Rule.
First
apply
Probability Principle
COMPLEMENT RULE: 𝑃 𝐸 = 1 βˆ’ 𝑃(𝐸 β€² )
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ π‘œπ‘›π‘’ π‘”π‘œπ‘‘
𝑃
=1βˆ’π‘ƒ
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
MATH 110 Sec 13.2 Operations with Events Practice Exercises
A sushi shop prepared 23 portions of seafood, 2 of which stayed out too
long and spoiled. If 18 of the 23 portions are served randomly, what is
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘
the probability that at least one customer
will receive spoiled food?
let’s
find 𝑃as a simplified fraction.
.
GiveSo,
final
answer
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘to be
(The
hint
on
the
previous
problem
tells
us
exactly
what
needs
As in the previous problem, it is easier to find 𝑃
first
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘
done here, but we will ignore that hint and just work it all out.)
and
thenthe
useBasic
the Complement
Rule.
First
apply
Probability Principle
β€²)
# π‘œπ‘“ π‘€π‘Žπ‘¦π‘ 
π‘‘π‘œ
𝑔𝑒𝑑
π‘›π‘œ
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
COMPLEMENT
RULE:
𝑃
𝐸
=
1
βˆ’
𝑃(𝐸
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
# π‘€π‘Žπ‘¦π‘ 
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘π‘œπ‘Ÿ π‘›π‘œπ‘‘)
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ π‘œπ‘›π‘’
π‘”π‘œπ‘‘ π‘‘π‘œ 𝑔𝑒𝑑 π‘Žπ‘›π‘¦ (π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
𝑃
=1βˆ’π‘ƒ
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
MATH 110 Sec 13.2 Operations with Events Practice Exercises
A sushi shop prepared 23 portions of seafood, 2 of which stayed out too
long and spoiled. If 18 of the 23 portions are served randomly, what is
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘
the probability that at least one customer
will receive spoiled food?
let’s
find 𝑃as a simplified fraction.
.
GiveSo,
final
answer
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘to be
(The
hint
on
the
previous
problem
tells
us
exactly
what
needs
In either
case,problem,
we are selecting
18 portions
As in the
previous
it is easier
to find but
𝑃 obviously order first
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
done here,
butThis
we tells
will ignore
hintcount
and just
work
it allπ‘“π‘œπ‘œπ‘‘
out.)
doesn’t
matter.
us thatthat
we will
using
combinations.
and then use the Complement Rule.
β€²)
# π‘œπ‘“ π‘€π‘Žπ‘¦π‘ 
π‘‘π‘œ
𝑔𝑒𝑑
π‘›π‘œ
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
COMPLEMENT
RULE:
𝑃
𝐸
=
1
βˆ’
𝑃(𝐸
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
# π‘€π‘Žπ‘¦π‘ 
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘π‘œπ‘Ÿ π‘›π‘œπ‘‘)
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ π‘œπ‘›π‘’
π‘”π‘œπ‘‘ π‘‘π‘œ 𝑔𝑒𝑑 π‘Žπ‘›π‘¦ (π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
𝑃
=1βˆ’π‘ƒ
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
MATH 110 Sec 13.2 Operations with Events Practice Exercises
A sushi shop prepared 23 portions of seafood, 2 of which stayed out too
long and spoiled. If 18 of the 23 portions are served randomly, what is
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘
the probability that at least one customer
will receive spoiled food?
let’s
find 𝑃as a simplified fraction.
.
GiveSo,
final
answer
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘to be
(The
hint
on
the
previous
problem
tells
us
exactly
what
needs
In either
case,problem,
we are selecting
18 portions
As in the
previous
it is easier
to find but
𝑃 obviously order first
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
done here,
butThis
we tells
will ignore
hintcount
and just
work
it allπ‘“π‘œπ‘œπ‘‘
out.)
doesn’t
matter.
us thatthat
we will
using
combinations.
and then use the Complement Rule.
β€²)
# π‘œπ‘“ π‘€π‘Žπ‘¦π‘ 
π‘‘π‘œ
𝑔𝑒𝑑
π‘›π‘œ
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
COMPLEMENT
RULE:
𝑃
𝐸
=
1
βˆ’
𝑃(𝐸
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
# π‘€π‘Žπ‘¦π‘ 
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘π‘œπ‘Ÿ π‘›π‘œπ‘‘)
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ π‘œπ‘›π‘’
π‘”π‘œπ‘‘ π‘‘π‘œ 𝑔𝑒𝑑 π‘Žπ‘›π‘¦ (π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
𝑃 π‘”π‘œπ‘‘
𝐢(21,18)= 1 βˆ’ 𝑃 π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
π‘›π‘œπ‘›π‘’
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
𝐢(23,18)
MATH 110 Sec 13.2 Operations with Events Practice Exercises
A sushi shop prepared 23 portions of seafood, 2 of which stayed out too
long and spoiled. If 18 of the 23 portions are served randomly, what is
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘
the probability that at least one customer
will receive spoiled food?
let’s
find 𝑃as a simplified fraction.
.
GiveSo,
final
answer
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘to be
(The
hint
on
the
previous
problem
tells
us
exactly
what
needs
In either
case,problem,
we are selecting
18 portions
As in the
previous
it is easier
to find but
𝑃 obviously order first
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
done here,
butThis
we tells
will ignore
hintcount
and just
work
it allπ‘“π‘œπ‘œπ‘‘
out.)
doesn’t
matter.
us thatthat
we will
using
combinations.
and then use the Complement Rule.
β€²)
# π‘œπ‘“ π‘€π‘Žπ‘¦π‘ 
π‘‘π‘œ
𝑔𝑒𝑑
π‘›π‘œ
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
COMPLEMENT
RULE:
𝑃
𝐸
=
1
βˆ’
𝑃(𝐸
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
# π‘€π‘Žπ‘¦π‘ 
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘π‘œπ‘Ÿ π‘›π‘œπ‘‘)
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ π‘œπ‘›π‘’
π‘”π‘œπ‘‘ π‘‘π‘œ 𝑔𝑒𝑑 π‘Žπ‘›π‘¦ (π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
𝑃 π‘”π‘œπ‘‘
𝐢(21,18)= 1 βˆ’ 𝑃 π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
π‘›π‘œπ‘›π‘’
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
𝐢(23,18)
MATH 110 Sec 13.2 Operations with Events Practice Exercises
A sushi shop prepared 23 portions of seafood, 2 of which stayed out too
long and spoiled. If 18 of the 23 portions are served randomly, what is
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘
the probability that at least one customer
will receive spoiled food?
let’s
find 𝑃as a simplified fraction.
.
GiveSo,
final
answer
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘to be
(The
hint
on
the
previous
problem
tells
us
exactly
what
needs
In either
case,problem,
we are selecting
18 portions
As in the
previous
it is easier
to find but
𝑃 obviously order first
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
done here,
butThis
we tells
will ignore
hintcount
and just
work
it allπ‘“π‘œπ‘œπ‘‘
out.)
doesn’t
matter.
us thatthat
we will
using
combinations.
and then use the Complement Rule.
β€²)
# π‘œπ‘“ π‘€π‘Žπ‘¦π‘ 
π‘‘π‘œ
𝑔𝑒𝑑
π‘›π‘œ
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
COMPLEMENT
RULE:
𝑃
𝐸
=
1
βˆ’
𝑃(𝐸
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
# π‘€π‘Žπ‘¦π‘ 
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘π‘œπ‘Ÿ π‘›π‘œπ‘‘)
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ π‘œπ‘›π‘’
π‘”π‘œπ‘‘ π‘‘π‘œ 𝑔𝑒𝑑 π‘Žπ‘›π‘¦ (π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
𝑃 π‘”π‘œπ‘‘
𝐢(21,18)= 1 βˆ’ 𝑃 π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
π‘›π‘œπ‘›π‘’
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
𝐢(23,18)
MATH
110 Sec
Operations
Practice
Exercises
To calculate
the13.2
number
of ways with
to getEvents
the portions,
spoiled
or not,
we can
choose ANY
of the 23 2
portions.
A sushi shop prepared
23 portions
of seafood,
of which stayed out too
long and spoiled. If 18 of the 23 portions are served randomly, what is
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘
the probability that at least one customer
will receive spoiled food?
let’s
find 𝑃as a simplified fraction.
.
GiveSo,
final
answer
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘to be
(The
hint
on
the
previous
problem
tells
us
exactly
what
needs
In either
case,problem,
we are selecting
18 portions
As in the
previous
it is easier
to find but
𝑃 obviously order first
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
done here,
butThis
we tells
will ignore
hintcount
and just
work
it allπ‘“π‘œπ‘œπ‘‘
out.)
doesn’t
matter.
us thatthat
we will
using
combinations.
and then use the Complement Rule.
β€²)
# π‘œπ‘“ π‘€π‘Žπ‘¦π‘ 
π‘‘π‘œ
𝑔𝑒𝑑
π‘›π‘œ
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
COMPLEMENT
RULE:
𝑃
𝐸
=
1
βˆ’
𝑃(𝐸
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
# π‘€π‘Žπ‘¦π‘ 
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘π‘œπ‘Ÿ π‘›π‘œπ‘‘)
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ π‘œπ‘›π‘’
π‘”π‘œπ‘‘ π‘‘π‘œ 𝑔𝑒𝑑 π‘Žπ‘›π‘¦ (π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
𝑃 π‘”π‘œπ‘‘
𝐢(21,18)= 1 βˆ’ 𝑃 π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
π‘›π‘œπ‘›π‘’
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
𝐢(23,18)
MATH
110 Sec
Operations
Practice
Exercises
To calculate
the13.2
number
of ways with
to getEvents
the portions,
spoiled
or not,
we can
choose ANY
of the 23 2
portions.
A sushi shop prepared
23 portions
of seafood,
of which stayed out too
long and spoiled. If 18 of the 23 portions are served randomly, what is
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘
the probability that at least one customer
will receive spoiled food?
let’s
find 𝑃as a simplified fraction.
.
GiveSo,
final
answer
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘to be
(The
hint
on
the
previous
problem
tells
us
exactly
what
needs
In either
case,problem,
we are selecting
18 portions
As in the
previous
it is easier
to find but
𝑃 obviously order first
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
done here,
butThis
we tells
will ignore
hintcount
and just
work
it allπ‘“π‘œπ‘œπ‘‘
out.)
doesn’t
matter.
us thatthat
we will
using
combinations.
and then use the Complement Rule.
β€²)
# π‘œπ‘“ π‘€π‘Žπ‘¦π‘ 
π‘‘π‘œ
𝑔𝑒𝑑
π‘›π‘œ
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
COMPLEMENT
RULE:
𝑃
𝐸
=
1
βˆ’
𝑃(𝐸
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
# π‘€π‘Žπ‘¦π‘ 
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘π‘œπ‘Ÿ π‘›π‘œπ‘‘)
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ π‘œπ‘›π‘’
π‘”π‘œπ‘‘ π‘‘π‘œ 𝑔𝑒𝑑 π‘Žπ‘›π‘¦ (π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
𝑃 π‘”π‘œπ‘‘
𝐢(21,18)= 1 βˆ’ 𝑃 π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
π‘›π‘œπ‘›π‘’
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
𝐢(23,18)
MATH
110 Sec
Operations
Practice
Exercises
To calculate
the13.2
number
of ways with
to getEvents
the portions,
spoiled
or not,
we can
choose ANY
of the 23 2
portions.
A sushi shop prepared
23 portions
of seafood,
of which stayed out too
long and spoiled. If 18 of the 23 portions are served randomly, what is
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘
the probability that at least one customer
will receive spoiled food?
let’s
find 𝑃as a simplified fraction.
.
GiveSo,
final
answer
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘to be
(The
hint
on
the
previous
problem
tells
us
exactly
what
needs
In either
case,problem,
we are selecting
18 portions
As in the
previous
it is easier
to find but
𝑃 obviously order first
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
done here,
butThis
we tells
will ignore
hintcount
and just
work
it allπ‘“π‘œπ‘œπ‘‘
out.)
doesn’t
matter.
us thatthat
we will
using
combinations.
and then use the Complement Rule.
β€²)
# π‘œπ‘“ π‘€π‘Žπ‘¦π‘ 
π‘‘π‘œ
𝑔𝑒𝑑
π‘›π‘œ
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
COMPLEMENT
RULE:
𝑃
𝐸
=
1
βˆ’
𝑃(𝐸
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
# π‘€π‘Žπ‘¦π‘ 
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘π‘œπ‘Ÿ π‘›π‘œπ‘‘)
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ π‘œπ‘›π‘’
π‘”π‘œπ‘‘ π‘‘π‘œ 𝑔𝑒𝑑 π‘Žπ‘›π‘¦ (π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
𝑃 π‘”π‘œπ‘‘
𝐢(21,18)= 1 βˆ’ 𝑃 π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
π‘›π‘œπ‘›π‘’
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
𝐢(23,18)
MATH 110 Sec 13.2 Operations with Events Practice Exercises
A sushi shop prepared 23 portions of seafood, 2 of which stayed out too
long and spoiled. If 18 of the 23 portions are served randomly, what is
the
probability
thatofatways
leasttoone
will
receive
food?
To find
the number
get customer
onlyπ‘›π‘œπ‘›π‘’
GOODπ‘”π‘œπ‘‘
portions
, wespoiled
can choose
let’s
find 𝑃asportions
. were spoiled).
only from among
the
21
GOOD
(because
2
GiveSo,
final
answer
a simplified
fraction.
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘to be
(The
hint
on
the
previous
problem
tells
us
exactly
what
needs
In either
case,problem,
we are selecting
18 portions
As in the
previous
it is easier
to find but
𝑃 obviously order first
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
done here,
butThis
we tells
will ignore
hintcount
and just
work
it allπ‘“π‘œπ‘œπ‘‘
out.)
doesn’t
matter.
us thatthat
we will
using
combinations.
and then use the Complement Rule.
β€²)
# π‘œπ‘“ π‘€π‘Žπ‘¦π‘ 
π‘‘π‘œ
𝑔𝑒𝑑
π‘›π‘œ
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
COMPLEMENT
RULE:
𝑃
𝐸
=
1
βˆ’
𝑃(𝐸
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
# π‘€π‘Žπ‘¦π‘ 
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘π‘œπ‘Ÿ π‘›π‘œπ‘‘)
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ π‘œπ‘›π‘’
π‘”π‘œπ‘‘ π‘‘π‘œ 𝑔𝑒𝑑 π‘Žπ‘›π‘¦ (π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
𝑃 π‘”π‘œπ‘‘
𝐢(21,18)= 1 βˆ’ 𝑃 π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
π‘›π‘œπ‘›π‘’
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
𝐢(23,18)
MATH 110 Sec 13.2 Operations with Events Practice Exercises
A sushi shop prepared 23 portions of seafood, 2 of which stayed out too
long and spoiled. If 18 of the 23 portions are served randomly, what is
the
probability
thatofatways
leasttoone
will
receive
food?
To find
the number
get customer
onlyπ‘›π‘œπ‘›π‘’
GOODπ‘”π‘œπ‘‘
portions
, wespoiled
can choose
let’s
find 𝑃asportions
. were spoiled).
only from among
the
21
GOOD
(because
2
GiveSo,
final
answer
a simplified
fraction.
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘to be
(The
hint
on
the
previous
problem
tells
us
exactly
what
needs
In either
case,problem,
we are selecting
18 portions
As in the
previous
it is easier
to find but
𝑃 obviously order first
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
done here,
butThis
we tells
will ignore
hintcount
and just
work
it allπ‘“π‘œπ‘œπ‘‘
out.)
doesn’t
matter.
us thatthat
we will
using
combinations.
and then use the Complement Rule.
β€²)
# π‘œπ‘“ π‘€π‘Žπ‘¦π‘ 
π‘‘π‘œ
𝑔𝑒𝑑
π‘›π‘œ
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
COMPLEMENT
RULE:
𝑃
𝐸
=
1
βˆ’
𝑃(𝐸
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
# π‘€π‘Žπ‘¦π‘ 
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘π‘œπ‘Ÿ π‘›π‘œπ‘‘)
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ π‘œπ‘›π‘’
π‘”π‘œπ‘‘ π‘‘π‘œ 𝑔𝑒𝑑 π‘Žπ‘›π‘¦ (π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
𝑃 π‘”π‘œπ‘‘
𝐢(21,18)= 1 βˆ’ 𝑃 π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
π‘›π‘œπ‘›π‘’
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
𝐢(23,18)
MATH 110 Sec 13.2 Operations with Events Practice Exercises
A sushi shop prepared 23 portions of seafood, 2 of which stayed out too
long and spoiled. If 18 of the 23 portions are served randomly, what is
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘
the probability that at least one customer
will receive spoiled food?
let’s
find 𝑃as a simplified fraction.
.
GiveSo,
final
answer
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘to be
(The
hint
on
the
previous
problem
tells
us
exactly
what
needs
In either
case,problem,
we are selecting
18 portions
As in the
previous
it is easier
to find but
𝑃 obviously order first
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
done here,
butThis
we tells
will ignore
hintcount
and just
work
it allπ‘“π‘œπ‘œπ‘‘
out.)
doesn’t
matter.
us thatthat
we will
using
combinations.
and then use the Complement Rule.
β€²)
# π‘œπ‘“ π‘€π‘Žπ‘¦π‘ 
π‘‘π‘œ
𝑔𝑒𝑑
π‘›π‘œ
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
COMPLEMENT
RULE:
𝑃
𝐸
=
1
βˆ’
𝑃(𝐸
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
# π‘€π‘Žπ‘¦π‘ 
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘π‘œπ‘Ÿ π‘›π‘œπ‘‘)
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ π‘œπ‘›π‘’
π‘”π‘œπ‘‘ π‘‘π‘œ 𝑔𝑒𝑑 π‘Žπ‘›π‘¦ (π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
𝑃 π‘”π‘œπ‘‘
𝐢(21,18)= 1 βˆ’ 𝑃 π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
π‘›π‘œπ‘›π‘’
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
𝐢(23,18)
MATH 110 Sec 13.2 Operations with Events Practice Exercises
A sushi shop prepared 23 portions of seafood, 2 of which stayed out too
long and spoiled. If 18 of the 23 portions are served randomly, what is
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘
the probability that at least one customer
will receive spoiled food?
let’s
find 𝑃as a simplified fraction.
.
GiveSo,
final
answer
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
Because
weon
must
the answer
astells
a simplified
fraction
(unliketoinbethe
(The hint
the give
previous
problem
us exactly
what needs
previous
problem),
weignore
will notthat
be able
to just
done
here,
but we will
hint and
justpunch
work the
it allentire
out.)
expression into the calculator and get the final answer.
and
then use
Rule. and reduce.
Instead, we must
calculator
thethe
topComplement
and bottom separately
β€²)
# π‘œπ‘“ π‘€π‘Žπ‘¦π‘ 
π‘‘π‘œ
𝑔𝑒𝑑
π‘›π‘œ
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
COMPLEMENT
RULE:
𝑃
𝐸
=
1
βˆ’
𝑃(𝐸
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
# π‘€π‘Žπ‘¦π‘ 
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘π‘œπ‘Ÿ π‘›π‘œπ‘‘)
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ π‘œπ‘›π‘’
π‘”π‘œπ‘‘ π‘‘π‘œ 𝑔𝑒𝑑 π‘Žπ‘›π‘¦ (π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
𝑃 π‘”π‘œπ‘‘
𝐢(21,18)= 1 βˆ’ 𝑃 π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
π‘›π‘œπ‘›π‘’
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
𝐢(23,18)
MATH 110 Sec 13.2 Operations with Events Practice Exercises
A sushi shop prepared 23 portions of seafood, 2 of which stayed out too
long and spoiled. If 18 of the 23 portions are served randomly, what is
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘
the probability that at least one customer
will receive spoiled food?
let’s
find 𝑃as a simplified fraction.
.
GiveSo,
final
answer
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
Because
weon
must
the answer
astells
a simplified
fraction
(unliketoinbethe
(The hint
the give
previous
problem
us exactly
what needs
previous
problem),
weignore
will notthat
be able
to just
done
here,
but we will
hint and
justpunch
work the
it allentire
out.)
expression into the calculator and get the final answer.
and
then use
Rule. and reduce.
Instead, we must
calculator
thethe
topComplement
and bottom separately
β€²)
# π‘œπ‘“ π‘€π‘Žπ‘¦π‘ 
π‘‘π‘œ
𝑔𝑒𝑑
π‘›π‘œ
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
COMPLEMENT
RULE:
𝑃
𝐸
=
1
βˆ’
𝑃(𝐸
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
# π‘€π‘Žπ‘¦π‘ 
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘π‘œπ‘Ÿ π‘›π‘œπ‘‘)
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ π‘œπ‘›π‘’
π‘”π‘œπ‘‘ π‘‘π‘œ 𝑔𝑒𝑑 π‘Žπ‘›π‘¦ (π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
𝑃 π‘”π‘œπ‘‘
𝐢(21,18)= 1 βˆ’ 𝑃 π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
π‘›π‘œπ‘›π‘’
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
𝐢(23,18)
Let’s
do the denominator
first.Practice Exercises
MATH 110 Sec 13.2
Operations
with Events
A sushi shop prepared 23 portions of seafood, 2 of which stayed out too
long and spoiled. If 18 of the 23 portions are served randomly, what is
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘
the probability that at least one customer
will receive spoiled food?
let’s
find 𝑃as a simplified fraction.
.
GiveSo,
final
answer
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
Because
weon
must
the answer
astells
a simplified
fraction
(unliketoinbethe
(The hint
the give
previous
problem
us exactly
what needs
previous
problem),
weignore
will notthat
be able
to just
done
here,
but we will
hint and
justpunch
work the
it allentire
out.)
expression into the calculator and get the final answer.
and
then use
Rule. and reduce.
Instead, we must
calculator
thethe
topComplement
and bottom separately
β€²)
# π‘œπ‘“ π‘€π‘Žπ‘¦π‘ 
π‘‘π‘œ
𝑔𝑒𝑑
π‘›π‘œ
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
COMPLEMENT
RULE:
𝑃
𝐸
=
1
βˆ’
𝑃(𝐸
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
# π‘€π‘Žπ‘¦π‘ 
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘π‘œπ‘Ÿ π‘›π‘œπ‘‘)
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ π‘œπ‘›π‘’
π‘”π‘œπ‘‘ π‘‘π‘œ 𝑔𝑒𝑑 π‘Žπ‘›π‘¦ (π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
𝑃 π‘”π‘œπ‘‘
𝐢(21,18)= 1 βˆ’ 𝑃 π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
π‘›π‘œπ‘›π‘’
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
𝐢(23,18)
Let’s
do the denominator
first.Practice Exercises
MATH 110 Sec 13.2
Operations
with Events
A sushi shop prepared 23 portions of seafood, 2 of which stayed out too
1 8 are served
33649
long and Keystrokes:
spoiled. If 182of3the 23 portions
randomly, what is
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘
the probability that at least one customer
will receive spoiled food?
let’s
find 𝑃as a simplified fraction.
.
GiveSo,
final
answer
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
Because
weon
must
the answer
astells
a simplified
fraction
(unliketoinbethe
(The hint
the give
previous
problem
us exactly
what needs
previous
problem),
weignore
will notthat
be able
to just
done
here,
but we will
hint and
justpunch
work the
it allentire
out.)
expression into the calculator and get the final answer.
and
then use
Rule. and reduce.
Instead, we must
calculator
thethe
topComplement
and bottom separately
β€²)
# π‘œπ‘“ π‘€π‘Žπ‘¦π‘ 
π‘‘π‘œ
𝑔𝑒𝑑
π‘›π‘œ
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
COMPLEMENT
RULE:
𝑃
𝐸
=
1
βˆ’
𝑃(𝐸
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
# π‘€π‘Žπ‘¦π‘ 
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘π‘œπ‘Ÿ π‘›π‘œπ‘‘)
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ π‘œπ‘›π‘’
π‘”π‘œπ‘‘ π‘‘π‘œ 𝑔𝑒𝑑 π‘Žπ‘›π‘¦ (π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
𝑃 π‘”π‘œπ‘‘
𝐢(21,18)= 1 βˆ’ 𝑃 π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
π‘›π‘œπ‘›π‘’
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
𝐢(23,18)
Let’s
do the denominator
first.Practice Exercises
MATH 110 Sec 13.2
Operations
with Events
A sushi shop prepared 23 portions of seafood, 2 of which stayed out too
1 8 are served
33649
long and Keystrokes:
spoiled. If 182of3the 23 portions
randomly, what is
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘
the probability that at least one customer
will receive spoiled food?
let’s
find 𝑃as a simplified fraction.
.
GiveSo,
final
answer
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
Because
weon
must
the answer
astells
a simplified
fraction
(unliketoinbethe
(The hint
the give
previous
problem
us exactly
what needs
previous
problem),
weignore
will notthat
be able
to just
done
here,
but we will
hint and
justpunch
work the
it allentire
out.)
expression into the calculator and get the final answer.
and
then use
Rule. and reduce.
Instead, we must
calculator
thethe
topComplement
and bottom separately
β€²)
# π‘œπ‘“ π‘€π‘Žπ‘¦π‘ 
π‘‘π‘œ
𝑔𝑒𝑑
π‘›π‘œ
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
COMPLEMENT
RULE:
𝑃
𝐸
=
1
βˆ’
𝑃(𝐸
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
# π‘€π‘Žπ‘¦π‘ 
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘π‘œπ‘Ÿ π‘›π‘œπ‘‘)
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ π‘œπ‘›π‘’
π‘”π‘œπ‘‘ π‘‘π‘œ 𝑔𝑒𝑑 π‘Žπ‘›π‘¦ (π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
𝑃 π‘”π‘œπ‘‘
𝐢(21,18)= 1 βˆ’ 𝑃 π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
π‘›π‘œπ‘›π‘’
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
𝐢(23,18)
Let’s
do the denominator
first.Practice Exercises
MATH 110 Sec 13.2
Operations
with Events
A sushi shop prepared 23 portions of seafood, 2 of which stayed out too
1 8 are served
33649
long and Keystrokes:
spoiled. If 182of3the 23 portions
randomly, what is
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘
the probability that at least one customer
will receive spoiled food?
let’s
find 𝑃as a simplified fraction.
.
GiveSo,
final
answer
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
Because
weon
must
the answer
astells
a simplified
fraction
(unliketoinbethe
(The hint
the give
previous
problem
us exactly
what needs
previous
problem),
weignore
will notthat
be able
to just
done
here,
but we will
hint and
justpunch
work the
it allentire
out.)
expression into the calculator and get the final answer.
and
then use
Rule. and reduce.
Instead, we must
calculator
thethe
topComplement
and bottom separately
β€²)
# π‘œπ‘“ π‘€π‘Žπ‘¦π‘ 
π‘‘π‘œ
𝑔𝑒𝑑
π‘›π‘œ
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
COMPLEMENT
RULE:
𝑃
𝐸
=
1
βˆ’
𝑃(𝐸
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
# π‘€π‘Žπ‘¦π‘ 
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘π‘œπ‘Ÿ π‘›π‘œπ‘‘)
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ π‘œπ‘›π‘’
π‘”π‘œπ‘‘ π‘‘π‘œ 𝑔𝑒𝑑 π‘Žπ‘›π‘¦ (π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
𝑃 π‘”π‘œπ‘‘
𝐢(21,18)= 1 βˆ’ 𝑃 π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
π‘›π‘œπ‘›π‘’
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
33649
Now do the with
numerator.
MATH 110 Sec 13.2 Operations
Events Practice Exercises
A sushi shop prepared 23 portions of seafood, 2 of which stayed out too
long and spoiled. If 18 of the 23 portions are served randomly, what is
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘
the probability that at least one customer
will receive spoiled food?
let’s
find 𝑃as a simplified fraction.
.
GiveSo,
final
answer
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
Because
weon
must
the answer
astells
a simplified
fraction
(unliketoinbethe
(The hint
the give
previous
problem
us exactly
what needs
previous
problem),
weignore
will notthat
be able
to just
done
here,
but we will
hint and
justpunch
work the
it allentire
out.)
expression into the calculator and get the final answer.
and
then use
Rule. and reduce.
Instead, we must
calculator
thethe
topComplement
and bottom separately
β€²)
# π‘œπ‘“ π‘€π‘Žπ‘¦π‘ 
π‘‘π‘œ
𝑔𝑒𝑑
π‘›π‘œ
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
COMPLEMENT
RULE:
𝑃
𝐸
=
1
βˆ’
𝑃(𝐸
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
# π‘€π‘Žπ‘¦π‘ 
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘π‘œπ‘Ÿ π‘›π‘œπ‘‘)
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ π‘œπ‘›π‘’
π‘”π‘œπ‘‘ π‘‘π‘œ 𝑔𝑒𝑑 π‘Žπ‘›π‘¦ (π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
𝑃 π‘”π‘œπ‘‘
𝐢(21,18)= 1 βˆ’ 𝑃 π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
π‘›π‘œπ‘›π‘’
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
33649
Now do the with
numerator.
MATH 110 Sec 13.2 Operations
Events Practice Exercises
A sushi shop prepared 23 portions of seafood, 2 of which stayed out too
1 8 are served
33649
long and Keystrokes:
spoiled. If 182of1the 23 portions
randomly, what is
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘
the probability that at least one customer
will receive spoiled food?
let’s
find 𝑃as a simplified fraction.
.
GiveSo,
final
answer
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
Because
weon
must
the answer
astells
a simplified
fraction
(unliketoinbethe
(The hint
the give
previous
problem
us exactly
what needs
previous
problem),
weignore
will notthat
be able
to just
done
here,
but we will
hint and
justpunch
work the
it allentire
out.)
expression into the calculator and get the final answer.
and
then use
Rule. and reduce.
Instead, we must
calculator
thethe
topComplement
and bottom separately
β€²)
# π‘œπ‘“ π‘€π‘Žπ‘¦π‘ 
π‘‘π‘œ
𝑔𝑒𝑑
π‘›π‘œ
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
COMPLEMENT
RULE:
𝑃
𝐸
=
1
βˆ’
𝑃(𝐸
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
# π‘€π‘Žπ‘¦π‘ 
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘π‘œπ‘Ÿ π‘›π‘œπ‘‘)
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ π‘œπ‘›π‘’
π‘”π‘œπ‘‘ π‘‘π‘œ 𝑔𝑒𝑑 π‘Žπ‘›π‘¦ (π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
𝑃 π‘”π‘œπ‘‘
𝐢(21,18)= 1 βˆ’ 𝑃 π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
π‘›π‘œπ‘›π‘’
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
33649
Now do the with
numerator.
MATH 110 Sec 13.2 Operations
Events Practice Exercises
A sushi shop prepared 23 portions of seafood, 2 of which stayed out too
1 8 are served1330
long and Keystrokes:
spoiled. If 182of1the 23 portions
randomly, what is
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘
the probability that at least one customer
will receive spoiled food?
let’s
find 𝑃as a simplified fraction.
.
GiveSo,
final
answer
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
Because
weon
must
the answer
astells
a simplified
fraction
(unliketoinbethe
(The hint
the give
previous
problem
us exactly
what needs
previous
problem),
weignore
will notthat
be able
to just
done
here,
but we will
hint and
justpunch
work the
it allentire
out.)
expression into the calculator and get the final answer.
and
then use
Rule. and reduce.
Instead, we must
calculator
thethe
topComplement
and bottom separately
β€²)
# π‘œπ‘“ π‘€π‘Žπ‘¦π‘ 
π‘‘π‘œ
𝑔𝑒𝑑
π‘›π‘œ
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
COMPLEMENT
RULE:
𝑃
𝐸
=
1
βˆ’
𝑃(𝐸
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
# π‘€π‘Žπ‘¦π‘ 
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘π‘œπ‘Ÿ π‘›π‘œπ‘‘)
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ π‘œπ‘›π‘’
π‘”π‘œπ‘‘ π‘‘π‘œ 𝑔𝑒𝑑 π‘Žπ‘›π‘¦ (π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
𝑃 π‘”π‘œπ‘‘
𝐢(21,18)= 1 βˆ’ 𝑃 π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
π‘›π‘œπ‘›π‘’
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
33649
Now do the with
numerator.
MATH 110 Sec 13.2 Operations
Events Practice Exercises
A sushi shop prepared 23 portions of seafood, 2 of which stayed out too
1 8 are served1330
long and Keystrokes:
spoiled. If 182of1the 23 portions
randomly, what is
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘
the probability that at least one customer
will receive spoiled food?
let’s
find 𝑃as a simplified fraction.
.
GiveSo,
final
answer
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
Because
weon
must
the answer
astells
a simplified
fraction
(unliketoinbethe
(The hint
the give
previous
problem
us exactly
what needs
previous
problem),
weignore
will notthat
be able
to just
done
here,
but we will
hint and
justpunch
work the
it allentire
out.)
expression into the calculator and get the final answer.
and
then use
Rule. and reduce.
Instead, we must
calculator
thethe
topComplement
and bottom separately
β€²)
# π‘œπ‘“ π‘€π‘Žπ‘¦π‘ 
π‘‘π‘œ
𝑔𝑒𝑑
π‘›π‘œ
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
COMPLEMENT
RULE:
𝑃
𝐸
=
1
βˆ’
𝑃(𝐸
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
# π‘€π‘Žπ‘¦π‘ 
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘π‘œπ‘Ÿ π‘›π‘œπ‘‘)
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ π‘œπ‘›π‘’
π‘”π‘œπ‘‘ π‘‘π‘œ 𝑔𝑒𝑑 π‘Žπ‘›π‘¦ (π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
𝑃 π‘”π‘œπ‘‘
𝐢(21,18)
π‘›π‘œπ‘›π‘’
1330 = 1 βˆ’ 𝑃 π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
33649
Now do the with
numerator.
MATH 110 Sec 13.2 Operations
Events Practice Exercises
A sushi shop prepared 23 portions of seafood, 2 of which stayed out too
1 8 are served1330
long and Keystrokes:
spoiled. If 182of1the 23 portions
randomly, what is
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘
the probability that at least one customer
will receive spoiled food?
let’s
find 𝑃as a simplified fraction.
.
GiveSo,
final
answer
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
Because
weon
must
the answer
astells
a simplified
fraction
(unliketoinbethe
(The hint
the give
previous
problem
us exactly
what needs
previous
problem),
weignore
will notthat
be able
to just
done
here,
but we will
hint and
justpunch
work the
it allentire
out.)
expression into the calculator and get the final answer.
and
then use
Rule. and reduce.
Instead, we must
calculator
thethe
topComplement
and bottom separately
β€²)
# π‘œπ‘“ π‘€π‘Žπ‘¦π‘ 
π‘‘π‘œ
𝑔𝑒𝑑
π‘›π‘œ
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
COMPLEMENT
RULE:
𝑃
𝐸
=
1
βˆ’
𝑃(𝐸
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
# π‘€π‘Žπ‘¦π‘ 
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘π‘œπ‘Ÿ π‘›π‘œπ‘‘)
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ π‘œπ‘›π‘’
π‘”π‘œπ‘‘ π‘‘π‘œ 𝑔𝑒𝑑 π‘Žπ‘›π‘¦ (π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
𝑃 π‘”π‘œπ‘‘
𝐢(21,18)
π‘›π‘œπ‘›π‘’
1330 = 1 βˆ’ 𝑃 π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
33649
we13.2
must
reduce thewith
fraction
to lowest
terms.
MATHFinally,
110 Sec
Operations
Events
Practice
Exercises
larger23
numbers
this
be time-consuming.
A sushi shopWith
prepared
portions
ofcan
seafood,
2 of which stayed out too
However,
here
strategy that
might help.
long and spoiled.
If 18 of
theis23a portions
are served
randomly, what is
back atthat
the at
combination
that
produced
the numerator…
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
theLook
probability
least one customer
will receive
spoiled food?
So,
let’s
findtest
𝑃as to
.
in this case,Give
𝐢(21,18)
and
see if the
fraction
will reduce
final
answer
a simplified
fraction.
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘
by the first number…21 in this case. If it works, great, if not, test
(The the
hintother
on thefactors
previous
tells usfrom
exactly
what
ofproblem
the number
small
to needs
large…to be
done here, but we
willthose
ignorewould
that hint
just
here
be and
7 and
3.work it all out.)
and then use the Complement Rule.
β€²)
# π‘œπ‘“ π‘€π‘Žπ‘¦π‘ 
π‘‘π‘œ
𝑔𝑒𝑑
π‘›π‘œ
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
COMPLEMENT
RULE:
𝑃
𝐸
=
1
βˆ’
𝑃(𝐸
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
# π‘€π‘Žπ‘¦π‘ 
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘π‘œπ‘Ÿ π‘›π‘œπ‘‘)
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ π‘œπ‘›π‘’
π‘”π‘œπ‘‘ π‘‘π‘œ 𝑔𝑒𝑑 π‘Žπ‘›π‘¦ (π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
𝑃 π‘”π‘œπ‘‘
𝐢(21,18)
π‘›π‘œπ‘›π‘’
1330 = 1 βˆ’ 𝑃 π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
33649
we13.2
must
reduce thewith
fraction
to lowest
terms.
MATHFinally,
110 Sec
Operations
Events
Practice
Exercises
larger23
numbers
this
be time-consuming.
A sushi shopWith
prepared
portions
ofcan
seafood,
2 of which stayed out too
However,
here
strategy that
might help.
long and spoiled.
If 18 of
theis23a portions
are served
randomly, what is
back atthat
the at
combination
that
produced
the numerator…
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
theLook
probability
least one customer
will receive
spoiled food?
So,
let’s
findtest
𝑃as to
.
in this case,Give
𝐢(21,18)
and
see if the
fraction
will reduce
final
answer
a simplified
fraction.
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘
by the first number…21 in this case. If it works, great, if not, test
(The the
hintother
on thefactors
previous
tells usfrom
exactly
what
ofproblem
the number
small
to needs
large…to be
done here, but we
willthose
ignorewould
that hint
just
here
be and
7 and
3.work it all out.)
and then use the Complement Rule.
β€²)
# π‘œπ‘“ π‘€π‘Žπ‘¦π‘ 
π‘‘π‘œ
𝑔𝑒𝑑
π‘›π‘œ
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
COMPLEMENT
RULE:
𝑃
𝐸
=
1
βˆ’
𝑃(𝐸
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
# π‘€π‘Žπ‘¦π‘ 
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘π‘œπ‘Ÿ π‘›π‘œπ‘‘)
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ π‘œπ‘›π‘’
π‘”π‘œπ‘‘ π‘‘π‘œ 𝑔𝑒𝑑 π‘Žπ‘›π‘¦ (π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
𝑃 π‘”π‘œπ‘‘
𝐢(21,18)
π‘›π‘œπ‘›π‘’
1330 = 1 βˆ’ 𝑃 π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
33649
we13.2
must
reduce thewith
fraction
to lowest
terms.
MATHFinally,
110 Sec
Operations
Events
Practice
Exercises
larger23
numbers
this
be time-consuming.
A sushi shopWith
prepared
portions
ofcan
seafood,
2 of which stayed out too
However,
here
strategy that
might help.
long and spoiled.
If 18 of
theis23a portions
are served
randomly, what is
back atthat
the at
combination
that
produced
the numerator…
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
theLook
probability
least one customer
will receive
spoiled food?
So,
let’s
findtest
𝑃as to
.
in this case,Give
𝐢(21,18)
and
see if the
fraction
will reduce
final
answer
a simplified
fraction.
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘
by the first number…21 in this case. If it works, great, if not, test
(The the
hintother
on thefactors
previous
tells usfrom
exactly
what
ofproblem
the number
small
to needs
large…to be
done here, but we
willthose
ignorewould
that hint
just
here
be and
7 and
3.work it all out.)
and then use the Complement Rule.
β€²)
# π‘œπ‘“ π‘€π‘Žπ‘¦π‘ 
π‘‘π‘œ
𝑔𝑒𝑑
π‘›π‘œ
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
COMPLEMENT
RULE:
𝑃
𝐸
=
1
βˆ’
𝑃(𝐸
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
# π‘€π‘Žπ‘¦π‘ 
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘π‘œπ‘Ÿ π‘›π‘œπ‘‘)
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ π‘œπ‘›π‘’
π‘”π‘œπ‘‘ π‘‘π‘œ 𝑔𝑒𝑑 π‘Žπ‘›π‘¦ (π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
𝑃 π‘”π‘œπ‘‘
𝐢(21,18)
π‘›π‘œπ‘›π‘’
1330 = 1 βˆ’ 𝑃 π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
33649
we13.2
must
reduce thewith
fraction
to lowest
terms.
MATHFinally,
110 Sec
Operations
Events
Practice
Exercises
larger23
numbers
this
be time-consuming.
A sushi shopWith
prepared
portions
ofcan
seafood,
2 of which stayed out too
However,
here
strategy that
might help.
long and spoiled.
If 18 of
theis23a portions
are served
randomly, what is
back atthat
the at
combination
that
produced
the numerator…
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
theLook
probability
least one customer
will receive
spoiled food?
So,
let’s
findtest
𝑃as to
.
in this case,Give
𝐢(21,18)
and
see if the
fraction
will reduce
final
answer
a simplified
fraction.
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘
by the first number…21 in this case. If it works, great, if not, test
(The the
hintother
on thefactors
previous
tells usfrom
exactly
what
ofproblem
the number
small
to needs
large…to be
done here, but we
willthose
ignorewould
that hint
just
here
be and
7 and
3.work it all out.)
and then use the Complement Rule.
β€²)
# π‘œπ‘“ π‘€π‘Žπ‘¦π‘ 
π‘‘π‘œ
𝑔𝑒𝑑
π‘›π‘œ
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
COMPLEMENT
RULE:
𝑃
𝐸
=
1
βˆ’
𝑃(𝐸
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
# π‘€π‘Žπ‘¦π‘ 
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘π‘œπ‘Ÿ π‘›π‘œπ‘‘)
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ π‘œπ‘›π‘’
π‘”π‘œπ‘‘ π‘‘π‘œ 𝑔𝑒𝑑 π‘Žπ‘›π‘¦ (π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
𝑃 π‘”π‘œπ‘‘
𝐢(21,18)
π‘›π‘œπ‘›π‘’
1330 = 1 βˆ’ 𝑃 π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
33649
we13.2
must
reduce thewith
fraction
to lowest
terms.
MATHFinally,
110 Sec
Operations
Events
Practice
Exercises
larger23
numbers
this
be time-consuming.
A sushi shopWith
prepared
portions
ofcan
seafood,
2 of which stayed out too
However,
here
strategy that
might help.
long and spoiled.
If 18 of
theis23a portions
are served
randomly, what is
back atthat
the at
combination
that
produced
the numerator…
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
theLook
probability
least one customer
will receive
spoiled food?
So,
let’s
findtest
𝑃as to
.
in this case,Give
𝐢(21,18)
and
see if the
fraction
will reduce
final
answer
a simplified
fraction.
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘
by the first number…21 in this case. If it works, great, if not, test
(The the
hintother
on thefactors
previous
tells usfrom
exactly
what
ofproblem
the number
small
to needs
large…to be
done here, but we
willthose
ignorewould
that hint
just
here
be and
7 and
3.work it all out.)
and then use the Complement Rule.
β€²)
# π‘œπ‘“ π‘€π‘Žπ‘¦π‘ 
π‘‘π‘œ
𝑔𝑒𝑑
π‘›π‘œ
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
COMPLEMENT
RULE:
𝑃
𝐸
=
1
βˆ’
𝑃(𝐸
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
# π‘€π‘Žπ‘¦π‘ 
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘π‘œπ‘Ÿ π‘›π‘œπ‘‘)
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ π‘œπ‘›π‘’
π‘”π‘œπ‘‘ π‘‘π‘œ 𝑔𝑒𝑑 π‘Žπ‘›π‘¦ (π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
𝑃 π‘”π‘œπ‘‘
𝐢(21,18)
π‘›π‘œπ‘›π‘’
1330 = 1 βˆ’ 𝑃 π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
33649
we13.2
must
reduce thewith
fraction
to lowest
terms.
MATHFinally,
110 Sec
Operations
Events
Practice
Exercises
larger23
numbers
this
be time-consuming.
A sushi shopWith
prepared
portions
ofcan
seafood,
2 of which stayed out too
However,
here
strategy that
might help.
long and spoiled.
If 18 of
theis23a portions
are served
randomly, what is
back atthat
the at
combination
that
produced
the numerator…
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
theLook
probability
least one customer
will receive
spoiled food?
So,
let’s
findtest
𝑃as to
.
in this case,Give
𝐢(21,18)
and
see if the
fraction
will reduce
final
answer
a simplified
fraction.
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘
by the first number…21 in this case. If it works, great, if not, test
(The the
hintother
on thefactors
previous
tells usfrom
exactly
what
ofproblem
the number
small
to needs
large…to be
done here, but we
willthose
ignorewould
that hint
just
here
be and
7 and
3.work it all out.)
and then use the Complement Rule.
β€²)
# π‘œπ‘“ π‘€π‘Žπ‘¦π‘ 
π‘‘π‘œ
𝑔𝑒𝑑
π‘›π‘œ
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
COMPLEMENT
RULE:
𝑃
𝐸
=
1
βˆ’
𝑃(𝐸
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
# π‘€π‘Žπ‘¦π‘ 
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘π‘œπ‘Ÿ π‘›π‘œπ‘‘)
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ π‘œπ‘›π‘’
π‘”π‘œπ‘‘ π‘‘π‘œ 𝑔𝑒𝑑 π‘Žπ‘›π‘¦ (π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
this strategy, see if 21
𝑃 π‘”π‘œπ‘‘
= 1 βˆ’Applying
𝑃
𝐢(21,18)
π‘›π‘œπ‘›π‘’
1330
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
divides
both top
and bottom
𝑃
= π‘“π‘œπ‘œπ‘‘
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
33649
evenly.divide 1330
Oops! 21 doesn’t
evenly so move on…trying 7 next.
we13.2
must
reduce thewith
fraction
to lowest
terms.
MATHFinally,
110 Sec
Operations
Events
Practice
Exercises
larger23
numbers
this
be time-consuming.
A sushi shopWith
prepared
portions
ofcan
seafood,
2 of which stayed out too
However,
here
strategy that
might help.
long and spoiled.
If 18 of
theis23a portions
are served
randomly, what is
back atthat
the at
combination
that
produced
the numerator…
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
theLook
probability
least one customer
will receive
spoiled food?
So,
let’s
findtest
𝑃as to
.
in this case,Give
𝐢(21,18)
and
see if the
fraction
will reduce
final
answer
a simplified
fraction.
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘
by the first number…21 in this case. If it works, great, if not, test
(The the
hintother
on thefactors
previous
tells usfrom
exactly
what
ofproblem
the number
small
to needs
large…to be
done here, but we
willthose
ignorewould
that hint
just
here
be and
7 and
3.work it all out.)
and then use the Complement Rule.
β€²)
# π‘œπ‘“ π‘€π‘Žπ‘¦π‘ 
π‘‘π‘œ
𝑔𝑒𝑑
π‘›π‘œ
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
COMPLEMENT
RULE:
𝑃
𝐸
=
1
βˆ’
𝑃(𝐸
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
# π‘€π‘Žπ‘¦π‘ 
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘π‘œπ‘Ÿ π‘›π‘œπ‘‘)
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ π‘œπ‘›π‘’
π‘”π‘œπ‘‘ π‘‘π‘œ 𝑔𝑒𝑑 π‘Žπ‘›π‘¦ (π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
this strategy, see if 21
𝑃 π‘”π‘œπ‘‘
= 1 βˆ’Applying
𝑃
𝐢(21,18)
π‘›π‘œπ‘›π‘’
1330
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
divides
both top
and bottom
𝑃
= π‘“π‘œπ‘œπ‘‘
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
33649
evenly.divide 1330
Oops! 21 doesn’t
evenly so move on…trying 7 next.
we13.2
must
reduce thewith
fraction
to lowest
terms.
MATHFinally,
110 Sec
Operations
Events
Practice
Exercises
larger23
numbers
this
be time-consuming.
A sushi shopWith
prepared
portions
ofcan
seafood,
2 of which stayed out too
However,
here
strategy that
might help.
long and spoiled.
If 18 of
theis23a portions
are served
randomly, what is
back atthat
the at
combination
that
produced
the numerator…
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
theLook
probability
least one customer
will receive
spoiled food?
So,
let’s
findtest
𝑃as to
.
in this case,Give
𝐢(21,18)
and
see if the
fraction
will reduce
final
answer
a simplified
fraction.
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘
by the first number…21 in this case. If it works, great, if not, test
(The the
hintother
on thefactors
previous
tells usfrom
exactly
what
ofproblem
the number
small
to needs
large…to be
done here, but we
willthose
ignorewould
that hint
just
here
be and
7 and
3.work it all out.)
and then use the Complement Rule.
β€²)
# π‘œπ‘“ π‘€π‘Žπ‘¦π‘ 
π‘‘π‘œ
𝑔𝑒𝑑
π‘›π‘œ
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
COMPLEMENT
RULE:
𝑃
𝐸
=
1
βˆ’
𝑃(𝐸
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
# π‘€π‘Žπ‘¦π‘ 
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘π‘œπ‘Ÿ π‘›π‘œπ‘‘)
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ π‘œπ‘›π‘’
π‘”π‘œπ‘‘ π‘‘π‘œ 𝑔𝑒𝑑 π‘Žπ‘›π‘¦ (π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
𝑃 π‘”π‘œπ‘‘
= 1 βˆ’ 𝑃 Does 7 divide both
𝐢(21,18)
π‘›π‘œπ‘›π‘’
1330
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘evenly?
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
top
and bottom
𝑃
= π‘“π‘œπ‘œπ‘‘
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
33649
YES!
(1330/7 = 190 & 33649/7 = 4807)
we13.2
must
reduce thewith
fraction
to lowest
terms.
MATHFinally,
110 Sec
Operations
Events
Practice
Exercises
larger23
numbers
this
be time-consuming.
A sushi shopWith
prepared
portions
ofcan
seafood,
2 of which stayed out too
However,
here
strategy that
might help.
long and spoiled.
If 18 of
theis23a portions
are served
randomly, what is
back atthat
the at
combination
that
produced
the numerator…
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
theLook
probability
least one customer
will receive
spoiled food?
So,
let’s
findtest
𝑃as to
.
in this case,Give
𝐢(21,18)
and
see if the
fraction
will reduce
final
answer
a simplified
fraction.
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘
by the first number…21 in this case. If it works, great, if not, test
(The the
hintother
on thefactors
previous
tells usfrom
exactly
what
ofproblem
the number
small
to needs
large…to be
done here, but we
willthose
ignorewould
that hint
just
here
be and
7 and
3.work it all out.)
and then use the Complement Rule.
β€²)
# π‘œπ‘“ π‘€π‘Žπ‘¦π‘ 
π‘‘π‘œ
𝑔𝑒𝑑
π‘›π‘œ
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
COMPLEMENT
RULE:
𝑃
𝐸
=
1
βˆ’
𝑃(𝐸
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
# π‘€π‘Žπ‘¦π‘ 
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘π‘œπ‘Ÿ π‘›π‘œπ‘‘)
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ π‘œπ‘›π‘’
π‘”π‘œπ‘‘ π‘‘π‘œ 𝑔𝑒𝑑 π‘Žπ‘›π‘¦ (π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
𝑃 π‘”π‘œπ‘‘
= 1 βˆ’ 𝑃 Does 7 divide both
𝐢(21,18)
π‘›π‘œπ‘›π‘’
1330
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘evenly?
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
top
and bottom
𝑃
= π‘“π‘œπ‘œπ‘‘
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
33649
YES!
(1330/7 = 190 & 33649/7 = 4807)
we13.2
must
reduce thewith
fraction
to lowest
terms.
MATHFinally,
110 Sec
Operations
Events
Practice
Exercises
larger23
numbers
this
be time-consuming.
A sushi shopWith
prepared
portions
ofcan
seafood,
2 of which stayed out too
However,
here
strategy that
might help.
long and spoiled.
If 18 of
theis23a portions
are served
randomly, what is
back atthat
the at
combination
that
produced
the numerator…
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
theLook
probability
least one customer
will receive
spoiled food?
So,
let’s
findtest
𝑃as to
.
in this case,Give
𝐢(21,18)
and
see if the
fraction
will reduce
final
answer
a simplified
fraction.
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘
by the first number…21 in this case. If it works, great, if not, test
(The the
hintother
on thefactors
previous
tells usfrom
exactly
what
ofproblem
the number
small
to needs
large…to be
done here, but we
willthose
ignorewould
that hint
just
here
be and
7 and
3.work it all out.)
and then use the Complement Rule.
β€²)
# π‘œπ‘“ π‘€π‘Žπ‘¦π‘ 
π‘‘π‘œ
𝑔𝑒𝑑
π‘›π‘œ
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
COMPLEMENT
RULE:
𝑃
𝐸
=
1
βˆ’
𝑃(𝐸
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
# π‘€π‘Žπ‘¦π‘ 
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘π‘œπ‘Ÿ π‘›π‘œπ‘‘)
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘
π‘œπ‘›π‘’
π‘”π‘œπ‘‘ π‘‘π‘œ 𝑔𝑒𝑑 π‘Žπ‘›π‘¦ (π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
7 × 190
𝑃 π‘”π‘œπ‘‘
= 1 βˆ’ 𝑃 Does 7 divide both
𝐢(21,18)
π‘›π‘œπ‘›π‘’
1330
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘evenly?
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
top
and bottom
𝑃
= π‘“π‘œπ‘œπ‘‘
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
33649
YES!
7 × 4807
(1330/7 = 190 & 33649/7 = 4807)
we13.2
must
reduce thewith
fraction
to lowest
terms.
MATHFinally,
110 Sec
Operations
Events
Practice
Exercises
larger23
numbers
this
be time-consuming.
A sushi shopWith
prepared
portions
ofcan
seafood,
2 of which stayed out too
However,
here
strategy that
might help.
long and spoiled.
If 18 of
theis23a portions
are served
randomly, what is
back atthat
the at
combination
that
produced
the numerator…
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
theLook
probability
least one customer
will receive
spoiled food?
So,
let’s
findtest
𝑃as to
.
in this case,Give
𝐢(21,18)
and
see if the
fraction
will reduce
final
answer
a simplified
fraction.
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘
by the first number…21 in this case. If it works, great, if not, test
(The the
hintother
on thefactors
previous
tells usfrom
exactly
what
ofproblem
the number
small
to needs
large…to be
done here, but we
willthose
ignorewould
that hint
just
here
be and
7 and
3.work it all out.)
and then use the Complement Rule.
β€²)
# π‘œπ‘“ π‘€π‘Žπ‘¦π‘ 
π‘‘π‘œ
𝑔𝑒𝑑
π‘›π‘œ
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
COMPLEMENT
RULE:
𝑃
𝐸
=
1
βˆ’
𝑃(𝐸
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
# π‘€π‘Žπ‘¦π‘ 
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘π‘œπ‘Ÿ π‘›π‘œπ‘‘)
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘
π‘œπ‘›π‘’
π‘”π‘œπ‘‘ π‘‘π‘œ 𝑔𝑒𝑑 π‘Žπ‘›π‘¦ (π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
7 × 190
𝑃 π‘”π‘œπ‘‘
= 1 βˆ’ 𝑃 Does 7 divide both
𝐢(21,18)
π‘›π‘œπ‘›π‘’
7
×
190
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘evenly?
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
top
and bottom
𝑃
= π‘“π‘œπ‘œπ‘‘
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
7 × 4807
YES!
7 × 4807
(1330/7 = 190 & 33649/7 = 4807)
we13.2
must
reduce thewith
fraction
to lowest
terms.
MATHFinally,
110 Sec
Operations
Events
Practice
Exercises
larger23
numbers
this
be time-consuming.
A sushi shopWith
prepared
portions
ofcan
seafood,
2 of which stayed out too
However,
here
strategy that
might help.
long and spoiled.
If 18 of
theis23a portions
are served
randomly, what is
back atthat
the at
combination
that
produced
the numerator…
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
theLook
probability
least one customer
will receive
spoiled food?
So,
let’s
findtest
𝑃as to
.
in this case,Give
𝐢(21,18)
and
see if the
fraction
will reduce
final
answer
a simplified
fraction.
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘
by the first number…21 in this case. If it works, great, if not, test
(The the
hintother
on thefactors
previous
tells usfrom
exactly
what
ofproblem
the number
small
to needs
large…to be
done here, but we
willthose
ignorewould
that hint
just
here
be and
7 and
3.work it all out.)
and then use the Complement Rule.
β€²)
# π‘œπ‘“ π‘€π‘Žπ‘¦π‘ 
π‘‘π‘œ
𝑔𝑒𝑑
π‘›π‘œ
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
COMPLEMENT
RULE:
𝑃
𝐸
=
1
βˆ’
𝑃(𝐸
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
# π‘€π‘Žπ‘¦π‘ 
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘π‘œπ‘Ÿ π‘›π‘œπ‘‘)
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ π‘œπ‘›π‘’
π‘”π‘œπ‘‘ π‘‘π‘œ 𝑔𝑒𝑑 π‘Žπ‘›π‘¦ (π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
𝑃 π‘”π‘œπ‘‘
= 1 βˆ’ 𝑃 Does 7 divide both
𝐢(21,18)
π‘›π‘œπ‘›π‘’
7
×
190
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘evenly?
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
top
and bottom
𝑃
= π‘“π‘œπ‘œπ‘‘
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
7 × 4807
YES!
(1330/7 = 190 & 33649/7 = 4807)
we13.2
must
reduce thewith
fraction
to lowest
terms.
MATHFinally,
110 Sec
Operations
Events
Practice
Exercises
larger23
numbers
this
be time-consuming.
A sushi shopWith
prepared
portions
ofcan
seafood,
2 of which stayed out too
However,
here
strategy that
might help.
long and spoiled.
If 18 of
theis23a portions
are served
randomly, what is
back atthat
the at
combination
that
produced
the numerator…
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
theLook
probability
least one customer
will receive
spoiled food?
So,
let’s
findtest
𝑃as to
.
in this case,Give
𝐢(21,18)
and
see if the
fraction
will reduce
final
answer
a simplified
fraction.
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘
by the first number…21 in this case. If it works, great, if not, test
(The the
hintother
on thefactors
previous
tells usfrom
exactly
what
ofproblem
the number
small
to needs
large…to be
done here, but we
willthose
ignorewould
that hint
just
here
be and
7 and
3.work it all out.)
and then use the Complement Rule.
β€²)
# π‘œπ‘“ π‘€π‘Žπ‘¦π‘ 
π‘‘π‘œ
𝑔𝑒𝑑
π‘›π‘œ
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
COMPLEMENT
RULE:
𝑃
𝐸
=
1
βˆ’
𝑃(𝐸
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
# π‘€π‘Žπ‘¦π‘ 
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘π‘œπ‘Ÿ π‘›π‘œπ‘‘)
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ π‘œπ‘›π‘’
π‘”π‘œπ‘‘ π‘‘π‘œ 𝑔𝑒𝑑 π‘Žπ‘›π‘¦ (π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
𝑃 π‘”π‘œπ‘‘
= 1 βˆ’ 𝑃 Does 7 divide both
𝐢(21,18)
π‘›π‘œπ‘›π‘’
7
×
190
\
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘evenly?
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
top
and bottom
𝑃
= π‘“π‘œπ‘œπ‘‘
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
\7 × 4807
YES!
(1330/7 = 190 & 33649/7 = 4807)
we13.2
must
reduce thewith
fraction
to lowest
terms.
MATHFinally,
110 Sec
Operations
Events
Practice
Exercises
larger23
numbers
this
be time-consuming.
A sushi shopWith
prepared
portions
ofcan
seafood,
2 of which stayed out too
However,
here
strategy that
might help.
long and spoiled.
If 18 of
theis23a portions
are served
randomly, what is
back atthat
the at
combination
that
produced
the numerator…
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
theLook
probability
least one customer
will receive
spoiled food?
So,
let’s
findtest
𝑃as to
.
in this case,Give
𝐢(21,18)
and
see if the
fraction
will reduce
final
answer
a simplified
fraction.
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘
by the first number…21 in this case. If it works, great, if not, test
(The the
hintother
on thefactors
previous
tells usfrom
exactly
what
ofproblem
the number
small
to needs
large…to be
done here, but we
willthose
ignorewould
that hint
just
here
be and
7 and
3.work it all out.)
and then use the Complement Rule.
β€²)
# π‘œπ‘“ π‘€π‘Žπ‘¦π‘ 
π‘‘π‘œ
𝑔𝑒𝑑
π‘›π‘œ
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
COMPLEMENT
RULE:
𝑃
𝐸
=
1
βˆ’
𝑃(𝐸
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
# π‘€π‘Žπ‘¦π‘ 
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘π‘œπ‘Ÿ π‘›π‘œπ‘‘)
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ π‘œπ‘›π‘’
π‘”π‘œπ‘‘ π‘‘π‘œ 𝑔𝑒𝑑 π‘Žπ‘›π‘¦ (π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
𝑃 π‘”π‘œπ‘‘
= 1 βˆ’ 𝑃 Does 7 divide both
𝐢(21,18)
π‘›π‘œπ‘›π‘’
190
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘evenly?
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
top
and bottom
𝑃
= π‘“π‘œπ‘œπ‘‘
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
4807
YES!
(1330/7 = 190 & 33649/7 = 4807)
we13.2
must
reduce thewith
fraction
to lowest
terms.
MATHFinally,
110 Sec
Operations
Events
Practice
Exercises
larger23
numbers
this
be time-consuming.
A sushi shopWith
prepared
portions
ofcan
seafood,
2 of which stayed out too
However,
here
strategy that
might help.
long and spoiled.
If 18 of
theis23a portions
are served
randomly, what is
back atthat
the at
combination
that
produced
the numerator…
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
theLook
probability
least one customer
will receive
spoiled food?
So,
let’s
findtest
𝑃as to
.
in this case,Give
𝐢(21,18)
and
see if the
fraction
will reduce
final
answer
a simplified
fraction.
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘
by the first number…21 in this case. If it works, great, if not, test
(The the
hintother
on thefactors
previous
tells usfrom
exactly
what
ofproblem
the number
small
to needs
large…to be
done here, but we
willthose
ignorewould
that hint
just
here
be and
7 and
3.work it all out.)
and then use the Complement Rule.
β€²)
# π‘œπ‘“ π‘€π‘Žπ‘¦π‘ 
π‘‘π‘œ
𝑔𝑒𝑑
π‘›π‘œ
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
COMPLEMENT
RULE:
𝑃
𝐸
=
1
βˆ’
𝑃(𝐸
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
# π‘€π‘Žπ‘¦π‘ 
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘π‘œπ‘Ÿ π‘›π‘œπ‘‘)
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ π‘œπ‘›π‘’
π‘”π‘œπ‘‘ π‘‘π‘œ 𝑔𝑒𝑑 π‘Žπ‘›π‘¦ (π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
𝑃 π‘”π‘œπ‘‘
= 1 βˆ’Now
𝑃 the top is small enough to
𝐢(21,18)
π‘›π‘œπ‘›π‘’
190
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
easily factor
intoπ‘“π‘œπ‘œπ‘‘
its prime factors.
𝑃
= π‘“π‘œπ‘œπ‘‘
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
4807
YES!
(1330/7 = 190 & 33649/7 = 4807)
we13.2
must
reduce thewith
fraction
to lowest
terms.
MATHFinally,
110 Sec
Operations
Events
Practice
Exercises
larger23
numbers
this
be time-consuming.
A sushi shopWith
prepared
portions
ofcan
seafood,
2 of which stayed out too
However,
here
strategy that
might help.
long and spoiled.
If 18 of
theis23a portions
are served
randomly, what is
back atthat
the at
combination
that
produced
the numerator…
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
theLook
probability
least one customer
will receive
spoiled food?
So,
let’s
findtest
𝑃as to
.
in this case,Give
𝐢(21,18)
and
see if the
fraction
will reduce
final
answer
a simplified
fraction.
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘
by the first number…21 in this case. If it works, great, if not, test
(The the
hintother
on thefactors
previous
tells usfrom
exactly
what
ofproblem
the number
small
to needs
large…to be
done here, but we
willthose
ignorewould
that hint
just
here
be and
7 and
3.work it all out.)
and then use the Complement Rule.
β€²)
# π‘œπ‘“ π‘€π‘Žπ‘¦π‘ 
π‘‘π‘œ
𝑔𝑒𝑑
π‘›π‘œ
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
COMPLEMENT
RULE:
𝑃
𝐸
=
1
βˆ’
𝑃(𝐸
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
# π‘€π‘Žπ‘¦π‘ 
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘π‘œπ‘Ÿ π‘›π‘œπ‘‘)
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ π‘œπ‘›π‘’
π‘”π‘œπ‘‘ π‘‘π‘œ 𝑔𝑒𝑑 π‘Žπ‘›π‘¦ (π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
𝑃 π‘”π‘œπ‘‘
= 1 βˆ’Now
𝑃 the top is small enough to
𝐢(21,18)
π‘›π‘œπ‘›π‘’
190
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
easily factor
intoπ‘“π‘œπ‘œπ‘‘
its prime factors.
𝑃
= π‘“π‘œπ‘œπ‘‘
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
4807
YES!
190 = 19 x 10 = 19 x 5 x 24807)
we13.2
must
reduce thewith
fraction
to lowest
terms.
MATHFinally,
110 Sec
Operations
Events
Practice
Exercises
larger23
numbers
this
be time-consuming.
A sushi shopWith
prepared
portions
ofcan
seafood,
2 of which stayed out too
However,
here
strategy that
might help.
long and spoiled.
If 18 of
theis23a portions
are served
randomly, what is
back atthat
the at
combination
that
produced
the numerator…
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
theLook
probability
least one customer
will receive
spoiled food?
So,
let’s
findtest
𝑃as to
.
in this case,Give
𝐢(21,18)
and
see if the
fraction
will reduce
final
answer
a simplified
fraction.
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘
by the first number…21 in this case. If it works, great, if not, test
(The the
hintother
on thefactors
previous
tells usfrom
exactly
what
ofproblem
the number
small
to needs
large…to be
done here, but we
willthose
ignorewould
that hint
just
here
be and
7 and
3.work it all out.)
and then use the Complement Rule.
β€²)
# π‘œπ‘“ π‘€π‘Žπ‘¦π‘ 
π‘‘π‘œ
𝑔𝑒𝑑
π‘›π‘œ
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
COMPLEMENT
RULE:
𝑃
𝐸
=
1
βˆ’
𝑃(𝐸
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
# π‘€π‘Žπ‘¦π‘ 
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘π‘œπ‘Ÿ π‘›π‘œπ‘‘)
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ π‘œπ‘›π‘’
π‘”π‘œπ‘‘ π‘‘π‘œ 𝑔𝑒𝑑 π‘Žπ‘›π‘¦ (π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
𝑃 π‘”π‘œπ‘‘
=
1 βˆ’Now
𝑃 the top is small enough to
𝐢(21,18)
π‘›π‘œπ‘›π‘’
19
×
5
×
2
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
easily factor
intoπ‘“π‘œπ‘œπ‘‘
its prime factors.
𝑃
= π‘“π‘œπ‘œπ‘‘
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
4807
YES!
190 = 19 x 10 = 19 x 5 x 24807)
we13.2
must
reduce thewith
fraction
to lowest
terms.
MATHFinally,
110 Sec
Operations
Events
Practice
Exercises
larger23
numbers
this
be time-consuming.
A sushi shopWith
prepared
portions
ofcan
seafood,
2 of which stayed out too
However,
here
strategy that
might help.
long and spoiled.
If 18 of
theis23a portions
are served
randomly, what is
back atthat
the at
combination
that
produced
the numerator…
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
theLook
probability
least one customer
will receive
spoiled food?
So,
let’s
findtest
𝑃as to
.
in this case,Give
𝐢(21,18)
and
see if the
fraction
will reduce
final
answer
a simplified
fraction.
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘
by the first number…21 in this case. If it works, great, if not, test
(The the
hintother
on thefactors
previous
tells usfrom
exactly
what
ofproblem
the number
small
to needs
large…to be
done here, but we
willthose
ignorewould
that hint
just
here
be and
7 and
3.work it all out.)
and then use the Complement Rule.
β€²)
# π‘œπ‘“ π‘€π‘Žπ‘¦π‘ 
π‘‘π‘œ
𝑔𝑒𝑑
π‘›π‘œ
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
COMPLEMENT
RULE:
𝑃
𝐸
=
1
βˆ’
𝑃(𝐸
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
# π‘€π‘Žπ‘¦π‘ 
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘π‘œπ‘Ÿ π‘›π‘œπ‘‘)
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ π‘œπ‘›π‘’
π‘”π‘œπ‘‘ π‘‘π‘œ 𝑔𝑒𝑑 π‘Žπ‘›π‘¦ (π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
𝑃 π‘”π‘œπ‘‘
=
1Now
βˆ’ 𝑃test each factor on top to see
𝐢(21,18)
π‘›π‘œπ‘›π‘’
19
×
5
×
2
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
if it evenly
divides
the bottom.
𝑃
= π‘“π‘œπ‘œπ‘‘
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
4807
YES!
)
we13.2
must
reduce thewith
fraction
to lowest
terms.
MATHFinally,
110 Sec
Operations
Events
Practice
Exercises
larger23
numbers
this
be time-consuming.
A sushi shopWith
prepared
portions
ofcan
seafood,
2 of which stayed out too
However,
here
strategy that
might help.
long and spoiled.
If 18 of
theis23a portions
are served
randomly, what is
back atthat
the at
combination
that
produced
the numerator…
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
theLook
probability
least one customer
will receive
spoiled food?
So,
let’s
findtest
𝑃as to
.
in this case,Give
𝐢(21,18)
and
see if the
fraction
will reduce
final
answer
a simplified
fraction.
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘
by the first number…21 in this case. If it works, great, if not, test
(The the
hintother
on thefactors
previous
tells usfrom
exactly
what
ofproblem
the number
small
to needs
large…to be
done here, but we
willthose
ignorewould
that hint
just
here
be and
7 and
3.work it all out.)
and then use the Complement Rule.
β€²)
# π‘œπ‘“ π‘€π‘Žπ‘¦π‘ 
π‘‘π‘œ
𝑔𝑒𝑑
π‘›π‘œ
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
COMPLEMENT
RULE:
𝑃
𝐸
=
1
βˆ’
𝑃(𝐸
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
# π‘€π‘Žπ‘¦π‘ 
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘π‘œπ‘Ÿ π‘›π‘œπ‘‘)
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ π‘œπ‘›π‘’
π‘”π‘œπ‘‘ π‘‘π‘œ 𝑔𝑒𝑑 π‘Žπ‘›π‘¦ (π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
𝑃 π‘”π‘œπ‘‘
=
1Now
βˆ’ 𝑃test each factor on top to see
𝐢(21,18)
π‘›π‘œπ‘›π‘’
19
×
5
×
2
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
if it evenly
divides
the bottom.
𝑃
= π‘“π‘œπ‘œπ‘‘
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
4807
YES!
4807/19 = 253 so 19 works!
we13.2
must
reduce thewith
fraction
to lowest
terms.
MATHFinally,
110 Sec
Operations
Events
Practice
Exercises
larger23
numbers
this
be time-consuming.
A sushi shopWith
prepared
portions
ofcan
seafood,
2 of which stayed out too
However,
here
strategy that
might help.
long and spoiled.
If 18 of
theis23a portions
are served
randomly, what is
back atthat
the at
combination
that
produced
the numerator…
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
theLook
probability
least one customer
will receive
spoiled food?
So,
let’s
findtest
𝑃as to
.
in this case,Give
𝐢(21,18)
and
see if the
fraction
will reduce
final
answer
a simplified
fraction.
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘
by the first number…21 in this case. If it works, great, if not, test
(The the
hintother
on thefactors
previous
tells usfrom
exactly
what
ofproblem
the number
small
to needs
large…to be
done here, but we
willthose
ignorewould
that hint
just
here
be and
7 and
3.work it all out.)
and then use the Complement Rule.
β€²)
# π‘œπ‘“ π‘€π‘Žπ‘¦π‘ 
π‘‘π‘œ
𝑔𝑒𝑑
π‘›π‘œ
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
COMPLEMENT
RULE:
𝑃
𝐸
=
1
βˆ’
𝑃(𝐸
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
# π‘€π‘Žπ‘¦π‘ 
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘π‘œπ‘Ÿ π‘›π‘œπ‘‘)
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ π‘œπ‘›π‘’
π‘”π‘œπ‘‘ π‘‘π‘œ 𝑔𝑒𝑑 π‘Žπ‘›π‘¦ (π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
𝑃 π‘”π‘œπ‘‘
=
1Now
βˆ’ 𝑃test each factor on top to see
𝐢(21,18)
π‘›π‘œπ‘›π‘’
19
×
5
×
2
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
if it evenly
divides
the bottom.
𝑃
= π‘“π‘œπ‘œπ‘‘
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
19 × 253
YES!
4807/19 = 253 so 19 works!
we13.2
must
reduce thewith
fraction
to lowest
terms.
MATHFinally,
110 Sec
Operations
Events
Practice
Exercises
larger23
numbers
this
be time-consuming.
A sushi shopWith
prepared
portions
ofcan
seafood,
2 of which stayed out too
However,
here
strategy that
might help.
long and spoiled.
If 18 of
theis23a portions
are served
randomly, what is
back atthat
the at
combination
that
produced
the numerator…
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
theLook
probability
least one customer
will receive
spoiled food?
So,
let’s
findtest
𝑃as to
.
in this case,Give
𝐢(21,18)
and
see if the
fraction
will reduce
final
answer
a simplified
fraction.
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘
by the first number…21 in this case. If it works, great, if not, test
(The the
hintother
on thefactors
previous
tells usfrom
exactly
what
ofproblem
the number
small
to needs
large…to be
done here, but we
willthose
ignorewould
that hint
just
here
be and
7 and
3.work it all out.)
and then use the Complement Rule.
β€²)
# π‘œπ‘“ π‘€π‘Žπ‘¦π‘ 
π‘‘π‘œ
𝑔𝑒𝑑
π‘›π‘œ
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
COMPLEMENT
RULE:
𝑃
𝐸
=
1
βˆ’
𝑃(𝐸
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
# π‘€π‘Žπ‘¦π‘ 
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘π‘œπ‘Ÿ π‘›π‘œπ‘‘)
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ π‘œπ‘›π‘’
π‘”π‘œπ‘‘ π‘‘π‘œ 𝑔𝑒𝑑 π‘Žπ‘›π‘¦ (π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
𝑃 π‘”π‘œπ‘‘
=
1Now
βˆ’ 𝑃test each factor on top to see
𝐢(21,18)
π‘›π‘œπ‘›π‘’
19
×
5
×
2
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
if it evenly
divides
the bottom.
𝑃
= π‘“π‘œπ‘œπ‘‘
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
19 × 253
YES!
4807/19 = 253 so 19 works!
we13.2
must
reduce thewith
fraction
to lowest
terms.
MATHFinally,
110 Sec
Operations
Events
Practice
Exercises
larger23
numbers
this
be time-consuming.
A sushi shopWith
prepared
portions
ofcan
seafood,
2 of which stayed out too
However,
here
strategy that
might help.
long and spoiled.
If 18 of
theis23a portions
are served
randomly, what is
back atthat
the at
combination
that
produced
the numerator…
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
theLook
probability
least one customer
will receive
spoiled food?
So,
let’s
findtest
𝑃as to
.
in this case,Give
𝐢(21,18)
and
see if the
fraction
will reduce
final
answer
a simplified
fraction.
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘
by the first number…21 in this case. If it works, great, if not, test
(The the
hintother
on thefactors
previous
tells usfrom
exactly
what
ofproblem
the number
small
to needs
large…to be
done here, but we
willthose
ignorewould
that hint
just
here
be and
7 and
3.work it all out.)
and then use the Complement Rule.
β€²)
# π‘œπ‘“ π‘€π‘Žπ‘¦π‘ 
π‘‘π‘œ
𝑔𝑒𝑑
π‘›π‘œ
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
COMPLEMENT
RULE:
𝑃
𝐸
=
1
βˆ’
𝑃(𝐸
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
# π‘€π‘Žπ‘¦π‘ 
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘π‘œπ‘Ÿ π‘›π‘œπ‘‘)
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ π‘œπ‘›π‘’
π‘”π‘œπ‘‘ π‘‘π‘œ 𝑔𝑒𝑑 π‘Žπ‘›π‘¦ (π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
𝑃 π‘”π‘œπ‘‘
= 1Now
βˆ’ 𝑃test each factor on top to see
𝐢(21,18)
π‘›π‘œπ‘›π‘’
5
×
2
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
if it evenly
divides
the bottom.
𝑃
= π‘“π‘œπ‘œπ‘‘
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
253
YES!
4807/19 = 253 so 19 works!
we13.2
must
reduce thewith
fraction
to lowest
terms.
MATHFinally,
110 Sec
Operations
Events
Practice
Exercises
larger23
numbers
this
be time-consuming.
A sushi shopWith
prepared
portions
ofcan
seafood,
2 of which stayed out too
However,
here
strategy that
might help.
long and spoiled.
If 18 of
theis23a portions
are served
randomly, what is
back atthat
the at
combination
that
produced
the numerator…
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
theLook
probability
least one customer
will receive
spoiled food?
So,
let’s
findtest
𝑃as to
.
in this case,Give
𝐢(21,18)
and
see if the
fraction
will reduce
final
answer
a simplified
fraction.
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘
by the first number…21 in this case. If it works, great, if not, test
(The the
hintother
on thefactors
previous
tells usfrom
exactly
what
ofproblem
the number
small
to needs
large…to be
done here, but we
willthose
ignorewould
that hint
just
here
be and
7 and
3.work it all out.)
and then use the Complement Rule.
β€²)
# π‘œπ‘“ π‘€π‘Žπ‘¦π‘ 
π‘‘π‘œ
𝑔𝑒𝑑
π‘›π‘œ
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
COMPLEMENT
RULE:
𝑃
𝐸
=
1
βˆ’
𝑃(𝐸
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
# π‘€π‘Žπ‘¦π‘ 
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘π‘œπ‘Ÿ π‘›π‘œπ‘‘)
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ π‘œπ‘›π‘’
π‘”π‘œπ‘‘ π‘‘π‘œ 𝑔𝑒𝑑 π‘Žπ‘›π‘¦ (π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
𝑃 π‘”π‘œπ‘‘
= 1Finally,
βˆ’ 𝑃 test 5 and 2 (on top) to see
𝐢(21,18)
π‘›π‘œπ‘›π‘’
5
×
2
π‘“π‘œπ‘œπ‘‘ the bottom.
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
if eitherπ‘ π‘π‘œπ‘–π‘™π‘’π‘‘
evenly divides
𝑃
= π‘“π‘œπ‘œπ‘‘
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
253
YES!
we13.2
must
reduce thewith
fraction
to lowest
terms.
MATHFinally,
110 Sec
Operations
Events
Practice
Exercises
larger23
numbers
this
be time-consuming.
A sushi shopWith
prepared
portions
ofcan
seafood,
2 of which stayed out too
However,
here
strategy that
might help.
long and spoiled.
If 18 of
theis23a portions
are served
randomly, what is
back atthat
the at
combination
that
produced
the numerator…
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
theLook
probability
least one customer
will receive
spoiled food?
So,
let’s
findtest
𝑃as to
.
in this case,Give
𝐢(21,18)
and
see if the
fraction
will reduce
final
answer
a simplified
fraction.
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘
by the first number…21 in this case. If it works, great, if not, test
(The the
hintother
on thefactors
previous
tells usfrom
exactly
what
ofproblem
the number
small
to needs
large…to be
done here, but we
willthose
ignorewould
that hint
just
here
be and
7 and
3.work it all out.)
and then use the Complement Rule.
β€²)
# π‘œπ‘“ π‘€π‘Žπ‘¦π‘ 
π‘‘π‘œ
𝑔𝑒𝑑
π‘›π‘œ
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
COMPLEMENT
RULE:
𝑃
𝐸
=
1
βˆ’
𝑃(𝐸
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
# π‘€π‘Žπ‘¦π‘ 
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘π‘œπ‘Ÿ π‘›π‘œπ‘‘)
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ π‘œπ‘›π‘’
π‘”π‘œπ‘‘ π‘‘π‘œ 𝑔𝑒𝑑 π‘Žπ‘›π‘¦ (π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
𝑃 π‘”π‘œπ‘‘
= 1Finally,
βˆ’ 𝑃 test 5 and 2 (on top) to see
𝐢(21,18)
π‘›π‘œπ‘›π‘’
5
×
2
π‘“π‘œπ‘œπ‘‘ the bottom.
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
if eitherπ‘ π‘π‘œπ‘–π‘™π‘’π‘‘
evenly divides
𝑃
= π‘“π‘œπ‘œπ‘‘
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
253
Neither 5 nor 2 evenly divides 253
thus we can’t simplify further.
we13.2
must
reduce thewith
fraction
to lowest
terms.
MATHFinally,
110 Sec
Operations
Events
Practice
Exercises
larger23
numbers
this
be time-consuming.
A sushi shopWith
prepared
portions
ofcan
seafood,
2 of which stayed out too
However,
here
strategy that
might help.
long and spoiled.
If 18 of
theis23a portions
are served
randomly, what is
back atthat
the at
combination
that
produced
the numerator…
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
theLook
probability
least one customer
will receive
spoiled food?
So,
let’s
findtest
𝑃as to
.
in this case,Give
𝐢(21,18)
and
see if the
fraction
will reduce
final
answer
a simplified
fraction.
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘
by the first number…21 in this case. If it works, great, if not, test
(The the
hintother
on thefactors
previous
tells usfrom
exactly
what
ofproblem
the number
small
to needs
large…to be
done here, but we
willthose
ignorewould
that hint
just
here
be and
7 and
3.work it all out.)
and then use the Complement Rule.
β€²)
# π‘œπ‘“ π‘€π‘Žπ‘¦π‘ 
π‘‘π‘œ
𝑔𝑒𝑑
π‘›π‘œ
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
COMPLEMENT
RULE:
𝑃
𝐸
=
1
βˆ’
𝑃(𝐸
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
# π‘€π‘Žπ‘¦π‘ 
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘π‘œπ‘Ÿ π‘›π‘œπ‘‘)
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ π‘œπ‘›π‘’
π‘”π‘œπ‘‘ π‘‘π‘œ 𝑔𝑒𝑑 π‘Žπ‘›π‘¦ (π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
𝑃 π‘”π‘œπ‘‘
= 1Finally,
βˆ’ 𝑃 test 5 and 2 (on top) to see
𝐢(21,18)
π‘›π‘œπ‘›π‘’
10
π‘“π‘œπ‘œπ‘‘ the bottom.
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
if eitherπ‘ π‘π‘œπ‘–π‘™π‘’π‘‘
evenly divides
𝑃
= π‘“π‘œπ‘œπ‘‘
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
253
Neither 5 nor 2 evenly divides 253
thus we can’t simplify further.
we13.2
must
reduce thewith
fraction
to lowest
terms.
MATHFinally,
110 Sec
Operations
Events
Practice
Exercises
larger23
numbers
this
be time-consuming.
A sushi shopWith
prepared
portions
ofcan
seafood,
2 of which stayed out too
However,
here
strategy that
might help.
long and spoiled.
If 18 of
theis23a portions
are served
randomly, what is
back atthat
the at
combination
that
produced
the numerator…
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
theLook
probability
least one customer
will receive
spoiled food?
So,
let’s
findtest
𝑃as to
.
in this case,Give
𝐢(21,18)
and
see if the
fraction
will reduce
final
answer
a simplified
fraction.
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘
by the first number…21 in this case. If it works, great, if not, test
(The the
hintother
on thefactors
previous
tells usfrom
exactly
what
ofproblem
the number
small
to needs
large…to be
done here, but we
willthose
ignorewould
that hint
just
here
be and
7 and
3.work it all out.)
and then use the Complement Rule.
β€²)
# π‘œπ‘“ π‘€π‘Žπ‘¦π‘ 
π‘‘π‘œ
𝑔𝑒𝑑
π‘›π‘œ
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘›π‘œπ‘›π‘’
π‘”π‘œπ‘‘
COMPLEMENT
RULE:
𝑃
𝐸
=
1
βˆ’
𝑃(𝐸
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
# π‘€π‘Žπ‘¦π‘ 
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘π‘œπ‘Ÿ π‘›π‘œπ‘‘)
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ π‘œπ‘›π‘’
π‘”π‘œπ‘‘ π‘‘π‘œ 𝑔𝑒𝑑 π‘Žπ‘›π‘¦ (π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
𝑃 π‘”π‘œπ‘‘
= 1Now
βˆ’ 𝑃 that we have this probability,
𝐢(21,18)
π‘›π‘œπ‘›π‘’
10
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
we can
go backπ‘“π‘œπ‘œπ‘‘
and plug it in to
𝑃
= π‘“π‘œπ‘œπ‘‘
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
253
get the final result that we were
looking for in the beginning.
MATH 110 Sec 13.2 Operations with Events Practice Exercises
# π‘œπ‘“ π‘€π‘Žπ‘¦π‘ 
π‘‘π‘œ 𝑔𝑒𝑑2π‘›π‘œ
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘›π‘œπ‘›π‘’prepared
π‘”π‘œπ‘‘
A sushi
shop
23
portions
of
seafood,
of
which stayed out too
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘
π‘€π‘Žπ‘¦π‘ 
π‘‘π‘œ 𝑔𝑒𝑑 π‘Žπ‘›π‘¦
π‘œπ‘Ÿ π‘›π‘œπ‘‘) what is
long and
spoiled.
If 18 of#the
23 portions
are(π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
served randomly,
Now that
wereceive
have this
probability,
the probability
least
will
spoiled
food?
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘that at 𝐢(21,18)
10one customer
we can go fraction.
back and plug it in to
𝑃
= answer as a simplified
Give
final
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
253
get the final result
thatπ‘”π‘œπ‘‘
we were
π‘›π‘œπ‘›π‘’
π‘›π‘œπ‘›π‘’
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘
π‘œπ‘›π‘’ π‘”π‘œπ‘‘
looking
for
inπ‘”π‘œπ‘‘
the beginning. first
As in the previous
problem,
it
is
easier
to
find
𝑃
𝑃
=1βˆ’π‘ƒ
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘ π‘“π‘œπ‘œπ‘‘
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
and then use the Complement Rule.
COMPLEMENT RULE: 𝑃 𝐸 = 1 βˆ’ 𝑃(𝐸 β€² )
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ π‘œπ‘›π‘’ π‘”π‘œπ‘‘
𝑃
=1βˆ’π‘ƒ
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
MATH 110 Sec 13.2 Operations with Events Practice Exercises
# π‘œπ‘“ π‘€π‘Žπ‘¦π‘ 
π‘‘π‘œ 𝑔𝑒𝑑2π‘›π‘œ
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘›π‘œπ‘›π‘’prepared
π‘”π‘œπ‘‘
A sushi
shop
23
portions
of
seafood,
of
which stayed out too
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘
π‘€π‘Žπ‘¦π‘ 
π‘‘π‘œ 𝑔𝑒𝑑 π‘Žπ‘›π‘¦
π‘œπ‘Ÿ π‘›π‘œπ‘‘) what is
long and
spoiled.
If 18 of#the
23 portions
are(π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
served randomly,
Now that
wereceive
have this
probability,
the probability
least
will
spoiled
food?
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘that at 𝐢(21,18)
10one customer
we can go fraction.
back and plug it in to
𝑃
= answer as a simplified
Give
final
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
253
get the final result
thatπ‘”π‘œπ‘‘
we were
π‘›π‘œπ‘›π‘’
π‘›π‘œπ‘›π‘’
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘
π‘œπ‘›π‘’ π‘”π‘œπ‘‘
looking
for
inπ‘”π‘œπ‘‘
the beginning. first
As in the previous
problem,
it
is
easier
to
find
𝑃
𝑃
=1βˆ’π‘ƒ
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘ π‘“π‘œπ‘œπ‘‘
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
and then use the Complement Rule.
COMPLEMENT RULE: 𝑃 𝐸 = 1 βˆ’ 𝑃(𝐸 β€² )
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ π‘œπ‘›π‘’ π‘”π‘œπ‘‘
𝑃
=1βˆ’π‘ƒ
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
MATH 110 Sec 13.2 Operations with Events Practice Exercises
# π‘œπ‘“ π‘€π‘Žπ‘¦π‘ 
π‘‘π‘œ 𝑔𝑒𝑑2π‘›π‘œ
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘›π‘œπ‘›π‘’prepared
π‘”π‘œπ‘‘
A sushi
shop
23
portions
of
seafood,
of
which stayed out too
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘
π‘€π‘Žπ‘¦π‘ 
π‘‘π‘œ 𝑔𝑒𝑑 π‘Žπ‘›π‘¦
π‘œπ‘Ÿ π‘›π‘œπ‘‘) what is
long and
spoiled.
If 18 of#the
23 portions
are(π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
served randomly,
Now that
wereceive
have this
probability,
the probability
least
will
spoiled
food?
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘that at 𝐢(21,18)
10one customer
we can go fraction.
back and plug it in to
𝑃
= answer as a simplified
Give
final
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
253
get the final result
thatπ‘”π‘œπ‘‘
we were
π‘›π‘œπ‘›π‘’
π‘›π‘œπ‘›π‘’
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘
π‘œπ‘›π‘’ π‘”π‘œπ‘‘
looking
for
inπ‘”π‘œπ‘‘
the beginning. first
As in the previous
problem,
it
is
easier
to
find
𝑃
𝑃
=1βˆ’π‘ƒ
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘ π‘“π‘œπ‘œπ‘‘
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
and then use the Complement Rule.
COMPLEMENT RULE: 𝑃 𝐸 = 1 βˆ’ 𝑃(𝐸 β€² )
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ π‘œπ‘›π‘’ π‘”π‘œπ‘‘
10π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘
𝑃
=1βˆ’π‘ƒ
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
253
MATH 110 Sec 13.2 Operations with Events Practice Exercises
# π‘œπ‘“ π‘€π‘Žπ‘¦π‘ 
π‘‘π‘œ 𝑔𝑒𝑑2π‘›π‘œ
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘›π‘œπ‘›π‘’prepared
π‘”π‘œπ‘‘
A sushi
shop
23
portions
of
seafood,
of
which stayed out too
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘
π‘€π‘Žπ‘¦π‘ 
π‘‘π‘œ 𝑔𝑒𝑑 π‘Žπ‘›π‘¦
π‘œπ‘Ÿ π‘›π‘œπ‘‘) what is
long and
spoiled.
If 18 of#the
23 portions
are(π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
served randomly,
Now that
wereceive
have this
probability,
the probability
least
will
spoiled
food?
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘that at 𝐢(21,18)
10one customer
we can go fraction.
back and plug it in to
𝑃
= answer as a simplified
Give
final
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
253
get the final result
thatπ‘”π‘œπ‘‘
we were
π‘›π‘œπ‘›π‘’
π‘›π‘œπ‘›π‘’
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘
π‘œπ‘›π‘’ π‘”π‘œπ‘‘
looking
for
inπ‘”π‘œπ‘‘
the beginning. first
As in the previous
problem,
it
is
easier
to
find
𝑃
𝑃
=1βˆ’π‘ƒ
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘ π‘“π‘œπ‘œπ‘‘
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
and then use the Complement Rule.
COMPLEMENT RULE: 𝑃 𝐸 = 1 βˆ’ 𝑃(𝐸 β€² )
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ π‘œπ‘›π‘’ π‘”π‘œπ‘‘
10π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘
𝑃
=1βˆ’π‘ƒ
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
253
MATH 110 Sec 13.2 Operations with Events Practice Exercises
# π‘œπ‘“ π‘€π‘Žπ‘¦π‘ 
π‘‘π‘œ 𝑔𝑒𝑑2π‘›π‘œ
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘›π‘œπ‘›π‘’prepared
π‘”π‘œπ‘‘
A sushi
shop
23
portions
of
seafood,
of
which stayed out too
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘
π‘€π‘Žπ‘¦π‘ 
π‘‘π‘œ 𝑔𝑒𝑑 π‘Žπ‘›π‘¦
π‘œπ‘Ÿ π‘›π‘œπ‘‘) what is
long and
spoiled.
If 18 of#the
23 portions
are(π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
served randomly,
Now that
wereceive
have this
probability,
the probability
least
will
spoiled
food?
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘that at 𝐢(21,18)
10one customer
we can go fraction.
back and plug it in to
𝑃
= answer as a simplified
Give
final
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
253
get the final result
thatπ‘”π‘œπ‘‘
we were
π‘›π‘œπ‘›π‘’
π‘›π‘œπ‘›π‘’
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘
π‘œπ‘›π‘’ π‘”π‘œπ‘‘
looking
for
inπ‘”π‘œπ‘‘
the beginning. first
As in the previous
problem,
it
is
easier
to
find
𝑃
𝑃
=1βˆ’π‘ƒ
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘ π‘“π‘œπ‘œπ‘‘
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
and then use the Complement Rule.
COMPLEMENT RULE: 𝑃 𝐸 = 1 βˆ’ 𝑃(𝐸 β€² )
π‘”π‘œπ‘‘ 10
243
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ π‘œπ‘›π‘’ π‘”π‘œπ‘‘
10π‘›π‘œπ‘›π‘’253
𝑃
=1βˆ’π‘ƒ
= π‘“π‘œπ‘œπ‘‘
βˆ’
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
253 253 253 253
MATH 110 Sec 13.2 Operations with Events Practice Exercises
# π‘œπ‘“ π‘€π‘Žπ‘¦π‘ 
π‘‘π‘œ 𝑔𝑒𝑑2π‘›π‘œ
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘›π‘œπ‘›π‘’prepared
π‘”π‘œπ‘‘
A sushi
shop
23
portions
of
seafood,
of
which stayed out too
𝑃
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘“π‘œπ‘œπ‘‘
π‘€π‘Žπ‘¦π‘ 
π‘‘π‘œ 𝑔𝑒𝑑 π‘Žπ‘›π‘¦
π‘œπ‘Ÿ π‘›π‘œπ‘‘) what is
long and
spoiled.
If 18 of#the
23 portions
are(π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
served randomly,
Now that
wereceive
have this
probability,
the probability
least
will
spoiled
food?
π‘›π‘œπ‘›π‘’ π‘”π‘œπ‘‘that at 𝐢(21,18)
10one customer
we can go fraction.
back and plug it in to
𝑃
= answer as a simplified
Give
final
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
253
get the final result
thatπ‘”π‘œπ‘‘
we were
π‘›π‘œπ‘›π‘’
π‘›π‘œπ‘›π‘’
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘
π‘œπ‘›π‘’ π‘”π‘œπ‘‘
looking
for
inπ‘”π‘œπ‘‘
the beginning. first
As in the previous
problem,
it
is
easier
to
find
𝑃
𝑃
=1βˆ’π‘ƒ
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘ π‘“π‘œπ‘œπ‘‘
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
and then use the Complement Rule.
COMPLEMENT RULE: 𝑃 𝐸 = 1 βˆ’ 𝑃(𝐸 β€² )
π‘”π‘œπ‘‘ 10
243
π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ π‘œπ‘›π‘’ π‘”π‘œπ‘‘
10π‘›π‘œπ‘›π‘’253
𝑃
=1βˆ’π‘ƒ
= π‘“π‘œπ‘œπ‘‘
βˆ’
=
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘
π‘ π‘π‘œπ‘–π‘™π‘’π‘‘ π‘“π‘œπ‘œπ‘‘
253 253 253 253
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