Download Probability of Mutually Exclusive and Inclusive Events

Geometry
Probability of Compound
Events
A single event is called a simple event. These
events have fairly simple probabilities.
Chance of rain
next Saturday in
Thousand Oaks
20%
P(Rain in T.O) = 0.2
Chance of rain
next Saturday in
Chicago
60%
P(Rain in Chicago) = 0.6
When flipping a coin twice (or flipping two coins at the same
time) having a result of Tails on the first coin toss and
Heads on the second is a simple event. However getting one
Tail and one Head is a compound event as there are two ways
in which that can happen. Look the Sample Space: {HH, TH,
HT, TT}
The weather in Thousand Oaks, CA,
doesn’t affect the weather in Chicago, IL.
These two events are called independent events because
the outcome of one doesn’t affect the outcome of the
other one.
Similarly the result of one coin toss does
not effect the result of another toss.
What if we looked at the probability of rain
occurring in both cities on Saturday?
When two or more simple events are combined, it
is considered a compound event. (Like the HT and
TH outcomes when tossing two coins form the
compound event of tossing one Head and one Tail)
Probability of Compound Events
If 2 events, A and B, are independent, then
the probability of both events occurring is
the product of the probability of A and the
probability of B.
P  A and B   P A  P B 
P  T.O. and Chicago   P T.O.   P  Chicago 

0.2

0.6
 0.12
The probability of rain occurring Saturday in
both T.O. and Chicago is 12%.
The event of it not raining in Chicago next Saturday is
a special kind of event called a Complement. In this
case it is the complement of it raining in Chicago next
Saturday. You can think of Complements as two events
that are opposites in terms of all the possibilities in
the sample space.
Example: When rolling a single die, the complement of
rolling a 3 is the event of rolling a 1, 2, 4, 5 or 6.
P  T.O. and not in Chicago   P T.O.   P not in Chicago 
 0.2
 0.08

0.4
If the chance of rain
in Chicago is 60%, then
the chance of it not
raining there is 40%.
The probability of rain occurring Saturday in
T.O. and not in Chicago is 8%.
Andrew is flying from Birmingham to Chicago. On the
first leg of the trip he has to fly from Birmingham to
Houston. In Houston he’ll change planes and head to
Chicago. Airline statistics report that the
Birmingham to Houston flight has a 90% on-time
record and the flight from Houston to Chicago has a
50% on-time record. Assuming that one flight’s ontime status is independent of another, what’s the
probability that both flights will be on time?
P  plane 1 and plane 2  P  plane 1   P  plane 2 
 0.9  0.5
 0.45
The probability of both flights being on time is 45%.
A die is rolled and a spinner like
the one shown on the right is
spun. Find each probability:
P  4 and A
 P  4   P A
1 1
 
6 4
1

24
P  even # and C 
A
B
D
C
 P  even #   P  C 
3 1
 
6 4
1

8
These are independent events. Multiply
the probabilities together to find the
probability of both occurring.
A die is rolled and a spinner like
the one shown on the right is
spun. Find each probability:
A
B
D
C
P  a number less than 5 and B, C, or D
 P  number less than 5   P  B, C, or D 
4 3
 
6 4
1

2
Find the probability that you’ll
roll a 6 and then a five when
you roll a die twice.
These are independent events. You will
multiply the probabilities together.
1
First roll : P  6  
6
1
Second roll : P  5  
6
P(6 and 5)
 P  6  P  5 
1 1
 
6 6
1

36
Round plastic chips numbered 1-15 are placed in a box. Chips
numbered 11-25 are placed in another box. A chip is
randomly drawn from the box on the left, then a second chip
is randomly drawn from the box on the right.
These are independent
events. Multiply the
probabilities together.
1-15
11-25
P both chips are greater than 14 
 P box 1   P box 2 
1 11


15 15
11

225
When the outcome of one event affects the
outcome of another event, the two events are
said to be dependent.
Probability of Dependent Events
If 2 events, A and B, are dependent, then the
probability of both events occurring is the
AandB  PAPB A
product of thePprobability
of A and the
probability of B given A occurs. This is called

a Conditional Probability.
P  A and B   P A  P B following A 
This can also be written using the formal notation for
conditional probability: P(A and B)=P(A)•P(B|A)
Here’s an example of two events that are dependent.
A bag contains 2 green, 9 brown, 7 yellow,
and 4 blue marbles. Once a marble is
selected, it is not replaced. Find the
probability of randomly drawing a brown,
then a yellow marble.
1st marble : P brown   9
22
2nd marble :
number of brown marbles
total # of marbles
7
P(yellow following brown) =
21
number of yellow marbles
total # of marbles remaining
So, putting all that all together, we see that:
P(brown and yellow) = P(brown)•P(yellow following brown)
9 7
3
=P(brown)•P(yellow|brown) 


22 21
22
Start again with a bag that contains 2
green, 9 brown, 7 yellow, and 4 blue
marbles. Again, once a marble is
selected, it is not replaced. Find the
probability of randomly drawing a green
marble, then a marble that’s not blue.
1st marble :
1
2
P  green  

11
22
2nd marble :
17
P(not blue|green) =
21
number of green marbles
total # of marbles
number of "not blue " marbles
total # of marbles remaining
1 17
17
P(green and not blue) 


11 21
231
Find the probability of randomly drawing
a yellow marble, a yellow marble, and a
blue marble. As before, marbles that are
drawn are not replaced.
1st marble :
7
P  yellow  
22
2nd marble :
6 2

P(yellow|yellow)=
21 7
3rd marble :
number of yellow marbles
total # of marbles
number of yellow marbles left
total # of marbles remaining
4 1

P(blue|yellow and yellow)=
20 5
number of blue marbles
total number of marbles remaining
P(yellow and yellow and blue)  7  2  1  1
22 7 5
55
At a school carnival, winners in a ring-toss game are
randomly given a prize from a bag that contains 4
sunglasses, 6 hairbrushes, and 5 key chains. Three
prizes are randomly chosen from the bag and not
replaced. Find the probability:
P(sunglasses and hairbrush and key chain)
=P(sunglasses)•P(hairbrush|sunglasses)•P(key chain|sunglasses and hairbrush)
3

4 6 5


15 14 13
3

4
91
7
Day 2
Probability of
Mutually Exclusive and Inclusive
Events
Mutually Exclusive Events
Mutually exclusive events, or disjoint
events, are events that cannot occur at
the same time.
For example, consider the events of rolling a 1 or a 3 on a
die? A die can’t show both a 1 and a 3 at the same time, so
the two events are considered mutually exclusive.
Probability of Mutually Exclusive Events
If 2 events, A and B, are mutually exclusive,
then the probability that either A or B
occurs is the sum of their probabilities.
P  A or B   P  A  P B

Find the probability
of rolling a 1 or a 3 on
a die.
P 1 or 3
 P 1   P  3 
1 1
 
6 6
1

3
These events are mutually
exclusive. The die can’t show
two different numbers at the
same time. This problem asks
what the probability is of one
or the other number showing.
In other words, an outcome of
1 or an outcome of 3 is a
desired outcome. The two
probabilities must be added
together to show that either
roll is acceptable.
Janet is going to an animal shelter to choose a new pet.
The shelter has 8 dogs, 7 cats, and 5 rabbits available
for adoption. If she randomly picks an animal to adopt,
what is the probability it will be a cat or a dog?
P  cat or dog 
 P  cat   P  dog 
7
8


20 20
15 3


20 4
The problem
asks for a
cat or a dog,
so either
outcome is
desired. The
two
probabilities
should be
added
together.
Bart randomly drew one
card from a standard deck
of cards. What is the
probability that the card
he drew was a club or a
diamond?
P  club or diamond 
 P  club   P  diamond 
13 13


52 52
26 1


52 2
Bart randomly drew one
card from a standard deck
of cards. What is the
probability that the card
he drew was a face card or
a 10?
P  face card or 10 
 P  face card   P 10 
12
4


52 52
16
4


52 13
Mutually Inclusive Events
Mutually inclusive events are events that
can occur at the same time.
For example, what is the probability of drawing either an
ace or a spade randomly from a deck of cards?
It’s possible for a card to be both an ace and a spade at the
same time. When we consider the probability of drawing
either an ace or a spade there are 4 cards in the deck that
are aces, 13 cards in the deck that are spades, and 1 card
that’s both an ace and a spade at the same time.
That one card should not be
counted twice!
Probability of Mutually Inclusive Events
If 2 events, A and B, are mutually inclusive, then
the probability that either A or B occurs is the sum
of their probabilities, decreased by the probability
of both occurring.
P  A or B   P A  P B   P A and B 
Let’s go back to the earlier question:
What is the probability of drawing
either an ace or a spade?
To figure this out you need to add the probability of
drawing an ace to the probability of drawing a spade, then
subtract out the probability of drawing a card that’s both
an ace and a spade at the same time.
P(ace or spade)  P(ace)  P(spade) P(ace and spade)
= P  ace   P  spade   P  ace of spades 
4 13
1



52 52 52
16
4


52 13
What is the probability of drawing
either a red card or an ace?
To figure this out you need to add the probability of
drawing a red card to the probability of drawing an ace,
then subtract out the probability of drawing a card that’s
both red and an ace at the same time.
P  red   P  ace   P red ace 
26 4
2



52 52 52
28

52
7

13
What is the probability of drawing a
face card or a spade?
P  face card or spade 
 P  face card   P  spade   P  spade face cards 
12 13 3



52 52 52

22
52
11

26
Suppose your dog had 9 puppies!
•
•
•
•
3 are brown females
2 are brown males
1 is a mixed color female
3 are mixed color males
If a puppy is randomly chosen from the litter,
what is the probability that it will be
male or be mixed color?
P  male or mixed 
 P  male   P  mixed   P  male and mixed 
5 4 3
  
9 9 9
6 2
 
9 3
The puppies that are
both male and mixed will
be counted twice if not
subtracted here.
In a bingo game, balls or tiles are
numbered from 1 to 75. The numbers
correspond to columns on a bingo card.
B
1-15
I
16-30
N
31-45
G
46-60
O
61-75
If a number is selected at random, what
is the probability it is
a multiple of 5 or in the “N” column?
What multiples of 5 appear in the “N” column?
35, 40, and 45
P  multiple of 5 or "N " column 
 P  multiple of 5   P "N" column   P  multiple of 5 and "N" column 
15 15 3



75 75 75

27
75
9

25
The numbers that are
both multiples of 5 and
appear in the “N”
column will be counted
twice if they’re not
subtracted out here.
P  even or "G " column 
 P  even   P  "G " column   P  evens in "G " column 

37 15 8


75 75 75
44

75
The numbers that are
both even and appear in
the “G” column will be
counted twice if
they’re not subtracted
out here.
Students are selected at random from a group of 12
boys and 12 girls. In that group there are 4 boys and 4
girls from each of 6th, 7th, and 8th grades.
Find the probability:
P  6th grade or girl 
 P  6th grade   P  girl   P  6th grade girl 

8 12 4


24 24 24

16
24
2

3
Students are selected at random from a group of 12
boys and 12 girls. In that group there are 4 boys and 4
girls from each of 6th, 7th, and 8th grades.
Find the probability:
P  male or not 8th 
 P  male   P not 8th   P  males not in 8th 

12 16 8


24 24 24

20
24
5

6
To sum it all up for today:
Mutually exclusive events are events that
cannot occur at the same time. You will see
the word “or” in the question.
Add the probabilities of mutually exclusive events
together to consider the probability that either one or
the other will occur.
Mutually inclusive events are events that
can occur at the same time. You will see
the word “or” in the question here too.
Add the probabilities of mutually inclusive events
together to consider the probability that either one or
the other will occur, but remember to subtract out the
events that overlap so they’re not counted twice.