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Conditional Probability
Example:
There are 5 red chip, 4 blue chips, and 6 white chips in a
basket. Two chips are randomly selected. Find the
probability that the second chip is red given that the first
chip is blue. (Assume that the first chip is not replaced.)
Because the first chip is selected and not replaced,
there are only 14 chips remaining.
5
 0.357
P (selecting a red chip | first chip is blue) 
14
Conditional Probability
Example:
100 college students were surveyed and asked how many hours
a week they spent studying. The results are in the table below.
Find the probability that a student spends more than 10 hours
studying given that the student is a male.
Male
Female
Total
Less
then 5
11
13
24
5 to 10
22
24
46
More
than 10
16
14
30
Total
49
51
100
The sample space consists of the 49 male students. Of
these 49, 16 spend more than 10 hours a week studying.
16
 0.327
P (more than 10 hours | male) 
49
Independent Events
Example:
Decide if the events are independent or dependent.
Selecting a diamond from a standard deck of cards (A), putting it
back in the deck, and then selecting a spade from the deck (B).
The occurrence of A does not affect the probability of B, so
the events are independent.
P (B A )  13  1 and P (B )  13  1 .
52
4
52
4
Multiplication Rule
Example:
Two cards are selected, without replacement, from a deck. Find
the probability of selecting a diamond, and then selecting a spade.
Because the card is not replaced, the events are dependent.
P (diamond and spade) = P (diamond) · P (spade |diamond).
13 13 169



 0.064
52 51 2652
Multiplication Rule
Example:
A die is rolled and two coins are tossed.
Find the probability of rolling a 5, and flipping two tails.
1
P (rolling a 5) =
.
6
Whether or not the roll is a 5, P (Tail ) = 1 ,
2
so the events are independent.
P (5 and T and T ) = P (5)· P (T )· P (T )
1 1 1
  
6 2 2
1

 0.042
24
Mutually Exclusive Events
Example:
Decide if the two events are mutually exclusive.
Event A: Roll a number less than 3 on a die. Event B: Roll
a 4 on a die.
A
B
1
2
4
These events cannot happen at the same time, so the events
are mutually exclusive.
Mutually Exclusive Events
Example:
Decide if the two events are mutually exclusive.
Event A: Select a Jack from a deck of cards.
Select a heart from a deck of cards.
A
J
J
J
Event B:
B
9 2
3 10
A 7
J
K
4
5 8
6
Q
Because the card can be a Jack and a heart at the same time,
the events are not mutually exclusive.
The Addition Rule
Example:
You roll a die. Find the probability that you roll a number less than 3
or a 4.
The events are mutually exclusive.
P (roll a number less than 3 or roll a 4)
= P (number is less than 3) + P (4)
2 1 3
    0.5
6 6 6
The Addition Rule
Example:
A card is randomly selected from a deck of cards. Find the probability
that the card is a Jack or the card is a heart.
The events are not mutually exclusive because the Jack of
hearts can occur in both events.
P (select a Jack or select a heart)
= P (Jack) + P (heart) – P (Jack of hearts)

4 13 1


52 52 52
16

52  0.308
The Addition Rule
Example:
100 college students were surveyed and asked how many hours a week
they spent studying. The results are in the table below. Find the
probability that a student spends between 5 and 10 hours or more than
10 hours studying.
Male
Female
Total
Less
then 5
11
13
24
5 to 10
22
24
46
More
than 10
16
14
30
Total
49
51
100
The events are mutually exclusive.
P (5 to10 hours or more than 10 ho P (5 to10) + P (10)
46 30
76
urs) =



 0.76
100 100 100
Review & Examples
Addition Rule
 P(A or B) = P(A) + P(B) – P(A and B)
↑
↑
either A or B or both A and B
A and B both occur at the same time
 If event A and B are disjoint or mutually exclusive (can not both
occur at the same time) then P(A and B) = 0, and P(A or B) = P(A)
+ P(B)
Review & Examples
General Multiplication Rule:
 P(A and B) = P(A) P(B/A) Or
P(B and A) = P(B) P(A/B)
Where P(B/A) represent the conditional probability of event B
given that event A has already occurred.
Independent Events: Two events A and B are independent if
knowing that one occurs does not change the probability that
the other occurs. If event A and B are independent then,
P(B/A) = P(B) and P(A and B) = P(A) P(B)
Examples
Example 1: Out of 36 people applying for the job, 20
are men and 16 are women. Eight of the men and 12 of
the women have Ph.D.’s. If one person is selected at
random for the first interview, find the probability that
the one chosen has a Ph. D.
 First organize the information:
PhD
Not PhD
Total
Men
8
12
20
Women
12
4
16
Total
20
16
36
Example Continued
a.
P(PhD) = 20/36
b.
the one chosen is a woman and has a Ph.D.
P(W and PhD) = 12/36
c.
the one chosen is a woman or has a Ph.D.
P(W or PhD) = P(W) + P(PhD) – P(W and PhD)
= 16/36 + 20/36 – 12/36
= 24/36 = 2/3
Examples
Example .
Of the 20 television programs to be aired this evening, Marc plans to
watch one, which he will pick at random by throwing a dart at TV
schedule. If 8 of the programs are educational, 9 are interesting, and
5 are both educational and interesting, find the probability that the
show he watches will have at least one of these attributes.
If E represent “educational” and I represent “interesting”, then
P(E) = 8/20, P(I) = 9/20, and P(E and I) = 5/20
8
9
5 12



P(E or I) =
20 20 20 20
Examples
3. The probability that a person selected at random did not
graduate from high school is 0.25. If three people are
selected at random, find the probability that
a.
all three do not have a high school diploma.
Since each person is independent ,
P(all three do not have a high school diploma) =(0.25)3=0.015625
b.
all three have a high school diploma.
P(all three have a high school diploma) = (1 − 0.25)= (0.75)3=0.42
c.
at least one has high school diploma.
P(at least one has HS diploma) = 1 – P(none have HS diploma)= 1 – (.25)3 =.9844
In general:
P(event happening at least once) = 1 – P(event does not happen)
Example Continued
Example: If a family has six children, find the probability that at
least one boy in the family?
There are 26 = 64 equally likely outcomes.
Since the complement of “at least one boy” is “all girls”
P(at least one boy) = 1 – P(all girls)
= 1 – 1/26 = 1 – 1/64 = 63/64
Examples
Example
The World Wide Insurance Company found that 53% of the residents
of a city had homeowner’s insurance with the company. Of these
clients, 27% also had car insurance with the company. If a resident is
selected at random, find the probability that the resident has both
homeowners and car insurance with the World Wide Insurance
Company.
Given: P(homeowner’s insurance) = 53%
P(car insurance / homeowner’s insurance) = 27%
P(homeowners and car insurance) = (.53) (.27) = .1431