Class 07 Conditional Probability David Housman for Math 323 Probability and Statistics

Conditional Probability Suppose an experiment is run which has a sample space S and an associated event A. S A B The probability of event A is Suppose that it is known that event B has occurred. The probability of event A is now P(A∩B) = P(A | B) P(B) David Housman for Math 323 Probability and Statistics

Student Gender and Hair Color One hundred students were asked their gender and hair color. The results are summarized in the table. Identify the population, sample, and variables. Suggest a statistic and a parameter. Suppose a student is selected at random from the sample (an experiment). Find the probabilities of the following events. student has red hair (1+5) / 100 = .06 student is male (27+32+1) / 100 = .60 student is male and has red hair 1 / 100 = .01 student is male or has red hair (100-18-17) / 100 = .65 student is male given that the student has red hair 1 / (1 + 5) = .167 student has red hair given that the student is male 1 / (27+32+1) = .017 David Housman for Math 323 Probability and Statistics

AIDS Example At the time, the test for the AIDS virus was 96% accurate. Ten thousand people considered to be at risk were tested for the AIDS virus. John Jones (not his real name) tested positive and two hours later committed suicide. In his last note, he explained that he saw no sense in continuing to live since he had a 96% chance of having AIDS. Is John Jones’ reasoning valid? 96 Missing information! test positive test negative have AIDS no AIDS 100 4 10000 396 test positive test negative 9900 9504 The number of persons testing positive is 96 + 396 = 492. The number of persons testing positive who have AIDS is 96. So, the probability that John Jones has AIDS is 96/492  20%. What are the implications for random workplace drug testing, mammograms, and early pregnancy test?

AIDS Example have AIDS no AIDS test positive test negative 10000 100 9900 96 4 396 9504 P(test negative | have AIDS) = 0.04 P(test positive | no AIDS) = 0.04 => P(test positive | have AIDS) = 0.96 => P(test negative | no AIDS) = 0.96 P(have AIDS) = 0.01 => P(no AIDS) = 0.99 P(test positive & have AIDS) = P(test positive | have AIDS) P(have AIDS) = (0.96) (0.01) = 0.0096 P(test positive & no AIDS) = P(test positive | no AIDS) P(no AIDS) = (0.04) (0.99) = 0.0396 P(test positive) = P(test positive & have AIDS) + P(test positive & no AIDS) = 0.0096 + 0.0396 = 0.0492 P(have AIDS | test positive) = P(test positive & have AIDS) / P(test positive) = 0.0096 / 0.0492 = 0.1951

David Housman for Math 323 Probability and Statistics Independence Roll a 4-sided die twice. A S (2,1) (1,2) (3,4) (4,3) (2,2) (1,3) (3,1) (4,4) (2,4) (1,1) (3,3) (4,2) (2,3) (1,4) (3,2) (4,1) Let A be “2 appears on the first roll.” C P(A) = 4 /16 = 1/4. Let B be “the sum is 5.” B P(A | B) = 1/4. Since P(A) = P(A | B), the event “2 appears on the first roll” is independent of the event “the sum is 5”. Let C be “the sum is 4.” P(A | C) = 1/3. Since P(A)  P(A | C), the event “2 appears on the first roll” is dependent upon the event “the sum is 4”. Event A is said to be independent of event B if P(A | B) = P(A). Otherwise, A is dependent on B. David Housman for Math 323 Probability and Statistics

Conditional Probability and Independence So, if knowing that B has occurred does not change the probability that A occurs, then knowing that A has occurred does not change the probability that B occurs. Also in this case, the probability that the events occur simultaneously is the product of the individual probabilities. We say events A and B are independent if P(AB) = P(A) P(B). David Housman for Math 323 Probability and Statistics

Conditional Probability and Independence So, if knowing that B has occurred does not change the probability that A occurs, then knowing that B has not occurred does not change the probability that A occurs. David Housman for Math 323 Probability and Statistics

Student Gender and Hair Color One hundred students were asked their gender and hair color. The results are summarized in the table. Suppose a student is selected at random from the sample (an experiment). Are the events “has red hair” and “is male” independent? P(red) = 6/100 P(red | male) = 1/60 No (find at least two other arguments) Are the events “has blonde hair” and “female” independent? P(blonde) = 45/100 = 9/20 P(blonde | female) = 18/40 = 9/20 Yes (find at least two other arguments) Two variables X and Y are independent if P(X = x & Y = y) = P(X = x) P(Y = y) for all x and y. Are gender and hair color independent? Change the numbers in the table so that the answer to the previous question changes.

David Housman for Math 323 Probability and Statistics Poker Suppose from a standard deck of cards that you have been dealt 4H, 5S, 6D, 6H, 9H. Given only this information, calculate the following probabilities. Obtain a heart flush by discarding 5S and 6D and drawing two new cards. There are 52 – 5 = 47 cards left from which we are drawing 2 There are 13 – 3 = 10 hearts left from which we need to draw 2 C(10, 2) / C(47, 2) = 0.0416 Obtain a straight by discarding 6H and 9H and drawing two new cards. Need to draw a 2 & 3 or a 3 & 7 or a 7 & 8; there are 4 of each rank to draw from 3 C(4, 1) C(4, 1) / C(47, 2) = 0.0444 Obtain a three of a kind, four of a kind, or full house by discarding 4H, 5S, and 9H and drawing three new cards. There are 52 – 5 = 47 cards left from which we are drawing 3 Need to draw one or two 6s from the remaining two 6s (C(2, 1) C(45, 2) + C(2, 2) C(45, 1)) / C(47, 3) = 0.1249 David Housman for Math 323 Probability and Statistics

More Probability Questions A bowl contains seven blue chips and three red chips. Two chips are to be drawn successively at random and without replacement. Find (1) the probability that the first draw is a red chip, and (2) the probability that the second draw is a blue chip. P(1st red) = 3/10 P(2nd blue) = P(1st red & 2nd blue) + P(1st blue & 2nd blue) = P(1st red) P(2nd blue | 1st red) + P(1st blue) P(2nd blue | 1st blue) = (3/10) (7/9) + (7/10) (6/9) = 7/10 = P(1st blue) From an ordinary deck of playing cards, cards are to be drawn successively at random and without replacement. Find the probability that the third spade appears on the sixth draw. P(3rd spade on 6th draw) = P(2 spades in 5 draws & spade on 6th draw) = P(2 spades in 5 draws) P(spade on 6th draw | 2 spades in 5 draws) = C(13, 2) C(39, 3) / C(52, 5) * (13 – 2) / (52 – 5) = 0.0642 David Housman for Math 323 Probability and Statistics

More Probability Questions A grade school boy has five blue and four white marbles in his left pocket and four blue and five white marbles in his right pocket. If he transfers one marble at random from his left to his right pocket, what is the probability of his then drawing a blue marble from his right pocket? P(pick B) = P(transfer B & pick B) + P(transfer W & pick B) = P(transfer B) P(pick B | transfer B) + P(transfer W) P(pick B | transfer W) = (5/9) (5/10) + (4/9)(4/10) = 41/90 A pair of four-sided dice are rolled. Find the probability of rolling a sum of 3 before rolling a sum of 5. x = P(sum 3 before sum 5) = P(1st sum 3) + P(1st sum not 3 or 5 & sum 3 before sum 5) = (2/16) + (10/16) x So, x = (2/16)(16/6) = 1/3 David Housman for Math 323 Probability and Statistics