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Transcript
Independence and
Conditional Probability
Warm-Up
•
If P(A) = 0.3 and P(B) =
0.4 and if A and B are
mutually exclusive
events, find:
a.
• a. 0.7
P( A)
b.
c.
P(B)
• b. 0.6
P ( A or B )
• c. 0.7
d. P( A and B)
• d. 0
Multiplication Rule –
Independent Events……
• When 2 events are independent, the
probability of both occurring is
P( A and B)  P( A)  P( B)
General Rule……
• “or” means to add
• “and” means to multiply (unless it is in
a contingency table and you can
actually see the intersection)
Example……
• If a coin is tossed
twice, find the
probability of
getting 2 heads.
• Answer:
P( H and H )  P( H )  P( H )
1 1 1
P( H and H )   
2 2 4
Example……
• A coin is flipped
and a die is rolled.
Find the
probability of
getting a head on
the coin and a 4 on
the die.
• Answer:
P( H and 4)  P( H )  P(4)
1 1 1
P( H and 4)   
2 6 12
Example……
• A card is drawn from
a deck and replaced;
then a 2nd card is
drawn. Find the
probability of getting
a queen and then an
ace.
• Answer:
P(Q and A)  P(Q)  P( A)
P(Q and A) 
4 4
1


52 52 169
Example……
• A box contains 3 red
balls, 2 blue balls, and
5 white balls. A ball is
selected and its color
noted. Then it is
replaced. A 2nd ball is
selected and its color
noted. Find the
probability of
a. Selecting 2 blue balls
b. Selecting a blue ball
and then a white ball
c. Selecting a red ball
and then a blue ball
Answers……
a. Selecting 2 blue balls
2 2
1
P( B and B)   
10 10 25
b. Selecting a blue ball
and then a white ball
2 5
1
P( B and W )   
10 10 10
c. Selecting a red ball
and then a blue ball
3 2 3
P( R and B)   
10 10 50
Example……
• A poll found that 46%
of Americans say they
suffer from stress.
If 3 people are
selected at random,
find the probability
that all three will say
they suffer from
stress.
• Answer:
P( S and S and S )  P( S )  P( S )  P( S )
P(Stress)  (0.46)  0.097
3
Dependent Events……
• When the outcome or occurrence of
the first event affects the outcome
or occurrence of the second event in
such a way that the probability is
changed.
Examples of Dependent
Events……
1. Draw a card from a deck. Do not
replace it and draw another card.
2. Having high grades and getting a
scholarship
3. Parking in a no parking zone and
getting a ticket
Multiplication Rule –
Dependent Events……
• When 2 events are dependent, the
probability of both occurring is
P( A and B)  P( A)  P( BlA )
• The slash reads:
“The probability that B occurs given
that A has already occurred.”
Example……
• 53% of residents had
homeowner’s
insurance. Of these,
27% also had car
insurance. If a
resident is selected at
random, find the prob.
That the resident has
both homeowner’s and
car insurance.
• Answer:
P( H and C )  P( H )  P(ClH )
P( H and C )  (.53)(.27)  .1431
Example……
• 3 cards are drawn from a deck and
NOT replaced. Find the following
probabilities.
a. Getting 3 jacks
b. Getting an ace, king, and queen
c. Getting a club, spade, and heart
d. Getting 3 clubs.
a. Getting 3 jacks……
4 3 2
1
P( J and J and J )   

 .000181
52 51 50 5525
b. Getting an ace, king,
queen……
4 4 4
8
P( A and K and Q) 
 

 .000483
52 51 50 16575
c. Getting a club, spade,
and heart……
13 13 13
169
P(C and S and H ) 
 

 .017
52 51 50 10200
d. Getting 3 clubs……
13 12 11 11
P(C and C and C ) 
 

or .013
52 51 50 850
Dependent Probability
Continued……Conditional
Warm Up……How Likely Are
You to Win the Lotto?
• Many states have lotteries. The biggest jackpot,
typically millions of dollars, usually comes from
the Lotto game. In Lotto South, available in
Georgia, Kentucky, and Virginia, six numbers are
randomly sampled without replacement from the
integers 1 to 49. The order of selection is not
important.
• Question: You buy a lottery ticket. What is
the probability that it is a winning ticket,
having the six numbers chosen?
• The probability of
winning is the
probability that
the 6 numbers
chosen are the six
that you have on
your ticket.
• Keep in mind that
the order does not
matter and that a
number cannot be
repeated after it
has been chosen.
• Find the
probability of
winning.
Answer……
6 5 4 3 2 1
P( All 6) 
    
49 48 47 46 45 44
720
P( All 6) 
 0.00000007
10,068,347,520
• This is about 1 chance in 14 million!
Insight……Provided by Wilson
and Crouch 2001, p. 200
• Let’s give this small number some
perspective. The chance of winning the
jackpot in Lotto South (0.00000007)
is……
• less than your chance of being hit by a
meteorite in the next year (0.0000004).
• less than your chance of dying in a
tornado (0.0000002).
• less than your chance of dying by a
lightning strike (0.00000016).
In other words……
• If you have money to spare, go ahead and
play the lottery, but understand why many
call it “sport for the mathematically
challenged.”
• By the way, the probability of winning the
Lotto South is also roughly the probability
that a person of average mortality will die
in the next 3 minutes!
•
Do you still want to play?
Back to Conditional
Probability - Remember……
P( A and B)  P( A)  PB A
• Algebraically change this so that it
is now in the form……
“Given”
P( A and B)
P  B A 
P( A)
P( A and B)
P A B  
P( B)
Example……
• In Rolling Acres Housing Plan, 42% of
the houses have a deck and a garage;
60% have a deck. Find the
probability that a home has a garage,
given that it has a deck.
Answer……
P( Deck and Garage)  .42
P( Deck )  .60
Find PG D
• Answer:
P(G and D)
PG D  
P ( D)
.42
PG D  
 .70
.6
Example……
• At an exclusive country club, 83% of
the members play bridge; 75% of the
members drink champagne given that
he or she plays bridge. Find the
probability that members drink
champagne and play bridge.
Answer……
• Answer:
P(bridge )  .83
P(C and B)
PC B  
P( B)
Pchamp bridge   .75
P ( B and C )
.75 
.83
Find P(champ and bridge )
P(C and B)  (.75)(. 83)  .62
Example……
• A recent survey
asked 100 people if
they thought women
in the armed forces
should be permitted
to participate in
combat. The results
are shown in the
table.
Yes
No
Total
Male
32
18
50
Female
8
42
50
Total
40
60
100
a. Find the probability that they
answered yes, given that they
were female.
Answer:
Yes
No
Total
Male
32
18
50
Female
8
42
50
Total
40
60
100
P(Y and F )
PY F  
P( F )
8
8
4
100
PY F  


or.16
50
50 25
100
b. Find the probability that they
were male, given that they
answered no.
• Answer:
Yes
No
Total
Male
32
18
50
Female
8
42
50
Total
40
60
100
P( M and No)
PM No 
P( No)
18
18 3
PM No  100 
 or.3
60
60 10
100