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.)

P (selecting a red chip | first chip is blue) 5 0.357

14

Because the first chip is selected and not replaced, there are only 14 chips remaining.

Conditional ProbabilityExample: 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.

P (more than 10 hours | male) 16 0.32749

The sample space consists of the 49 male students. Of these 49, 16 spend more than 10 hours a week studying.

Less then 5

5 to 10More

than 10Total

Male 11 22 16 49Female

13 24 14 51

Total 24 46 30 100

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).

13 1 13 1( ) and ( ) .52 4 52 4

P B A P B

The occurrence of A does not affect the probability of B, so the events are independent.

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 0.06452 51 2652

Example:A die is rolled and two coins are tossed. Find the probability of rolling a 5, and flipping two tails.

Multiplication Rule

P (rolling a 5) = 1.6

Whether or not the roll is a 5, P (Tail ) = so the events are independent.

1,2

P (5 and T and T ) = P (5)· P (T )· P (T ) 1 1 16 2 2

124

0.042

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.

These events cannot happen at the same time, so the events are mutually exclusive.

A

12

B

4

A

J

J

J

J

Mutually Exclusive Events

Example:Decide if the two events are mutually exclusive.

Event A: Select a Jack from a deck of cards. Event B:

Select a heart from a deck of cards.

Because the card can be a Jack and a heart at the same time, the events are not mutually exclusive.

B23

45

6

AK

Q

109

8

7

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 0.56 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 13 152 52 52

1652

0.308

The Addition RuleExample: 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.

P (5 to10) + P (10) 46 30

100 100

The events are mutually exclusive.

Less then 5

5 to 10More

than 10Total

Male 11 22 16 49Female 13 24 14 51Total 24 46 30 100

P (5 to10 hours or more than 10 hours) = 76 0.76

100

Review & Examples

Addition RuleP(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:Men Women Total

PhD 8 12 20

Not PhD 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

P(E or I) = 20

12

20

5

20

9

20

8

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