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

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. Less then 5 5 to 10 More than 10 Total Male 11 22 16 49 Female 13 24 14 51 46 30 100 The sample space consists of the 49 male students. Of these 49, 16 spend more than 10 hours a week studying. P (more than 10 hours | male)

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.

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).

Multiplication Rule Example: A die is rolled and two coins are tossed. Find the probability of rolling a 5, and flipping two tails. P (rolling a 5) = Whether or not the roll is a 5, P (Tail ) = so the events are independent. P (5 and T and T ) = P (5)· P (T )· P (T )

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 1 2 B 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. Event B: Select a heart from a deck of cards. A J J J J B 2 3 4 5 6 A K Q 10 9 8 7 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)

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)

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. Less then 5 5 to 10 More than 10 Total Male 11 22 16 49 Female 13 24 14 51 46 30 100 The events are mutually exclusive. P (5 to10 hours or more than 10 hours) = P (5 to10) + P (10)

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: Men Women Total PhD 8 12 20 Not PhD 4 16 36

Example Continued = 16/36 + 20/36 – 12/36 = 24/36 = 2/3 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

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