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Solutions to Problem Set #4 Section 4.1 18. A doctor assumes that a patient has one of three diseases d1 , d2 or d3 . Before any test, he assumes an equal probability for each disease. He carries out a test that will be positive with probability .8 if the patient has d1 , .6 if the patient has d2 and .4 if the patient has d3 . Given that the outcome of the test was positive, what probabilities should the doctor now assign to the three possible diseases? We are given that P (d1 ) = P (d2 ) = P (d3 ) = 31 . We’re also told that P (+ | d1 ) = .8, P (+ | d2 ) = .6 and P (+ | d3 ) = .4. In order to use Bayes’ Theorem, we need to compute: 1 1 1 + .6 + .4 . P (+) = P (+ | d1 ) · d1 + P (+ | d2 ) · d2 + P (+ | d3 ) · d3 = .8 3 3 3 Thus, by Bayes’ Theorem, we have: .8( 13 ) P (+ | d1 ) · P (d1 ) P (d1 | +) = = = .4444. P (+) .6 Using Bayes’ Theorem for the other two cases, we obtain P (d2 | +) = and P (d3 | +) = .6( 13 ) 1 = .6 3 .4( 13 ) = .2222. .6 22. One coin in a collection of 65 has two heads. The rest are fair. If a coin, chosen at random from the lot and then tossed, turns up heads 6 times in a row, what is the probability that it is the two-headed coin? Let E be the event “you pick the unfair coin.” Let F be the event: “you get 6 heads in a row.” Note that E ⊂ F . Let ω1 be the outcome “a fair coin is drawn, and flips 6 heads in a row.” Let ω2 be the outcome “the unfair coin is drawn, and (obviously) flips 6 heads 1 1 in a row.” Then E = {ω2 } and F = {ω1 , ω2 }. The probability of ω1 is 64 · 64 = 65 . The 65 1 probability of ω2 is 65 , since it is the same probability of just choosing the unfair coin to ) begin with. Thus, P (E | F ) = P P(E∩F = 12 . (F ) Section 6.1 2. A card is drawn from a deck of playing cards. If it is red, the player wins 1 dollar; if it is black, the player loses 2 dollars. Find the expected value of the game. Let X be the amount of money gained. Then E(X) = 12 (1) − 21 (2) = − 12 . 1 Solutions to Problem Set #4 8. A royal family has children until it has a boy or until it has three children, whichever comes first. Assume that each child is a boy with probability 12 . Find the expected number of boys and the expected number of girls in this royal family. Let X be the number of boys and Y be the number of girls. Then, E(X) = 1 1 1 7 + + = . 2 4 8 8 On the other hand, 1 1 7 1 E(Y ) = (3) + (2) + (1) = . 8 8 4 8 This calculation is done by observing that with three girls, you have probability 18 . In order to have 2 girls and 1 boy, you would need the order to be GGB, hence the probability is 18 . In order to have 1 girl and 1 boy, the order must be GB, hence the probability is 14 . 2