Problem Set 9 (a) Suppose your friend says, "I have the ace of spades. " What is the probablity that she has another ace? Solution. The sample space for this experient is the set of all 5-card hands. All outcomes are equally likely, so the probability of each outcome is 1/5). Let S be the event that your friend has the ace of spades, and let A be the event that your friend has another ace. Our objective is to compute Pr(A|S)Pr(A∩S) The number of hands containing the ace of spades is equal to the number of ways to select 4 of the remaining 51 cards. Therefore P(9= The number of hands containing the ace of spades and at least one more ace is 1A∩S Here the first term counts the number of hands with one additional ace, since there are ways to choose the extra ace and(3)ways to choose the other cards.Sim ilarly, the second term counts hands with two additional aces and the last term counts hands with all three remaining aces. In probability terms, we have: P:(A∩S=()+(2(2)+((2 Substituting these results into our original equation gives the solution Pr(4S)=3+23101=02214 (b) Suppose your friend says, I have an ace. What is the probability that she has another ace Solution. The sample space and outcome probabilities are the same as before. Let L be the event that your friend has at least one ace, and m be the event that your friend has more than one ace. Our goal it to compute M∩L)Pr(M) Pr(L The second equality holds because your friend surely at least one ace if she has mor than one; that is, M CL. The probability that your friend has at least one ace is Pr(L) )(4)+(2)(3)+(3)(2)+()(1)� � � � � � � �� � � �� � � �� � � � � � � �� � � �� � � �� � � � � �� � � �� � � �� � = � � = � �� � � �� � � �� � � �� � � � 6 52 Problem Set 9 (a) Suppose your friend says, “I have the ace of spades.” What is the probablity that she has another ace? Solution. The sample space for this experient is the set of all 5­card hands. All outcomes are equally likely, so the probability of each outcome is 1/ . Let S be 5 the event that your friend has the ace of spades, and let A be the event that your friend has another ace. Our objective is to compute: Pr (A | S) = Pr (A ∩ S) Pr (S) The number of hands containing the ace of spades is equal to the number of ways to select 4 of the remaining 51 cards. Therefore: 51 4 Pr (S) = 52 5 The number of hands containing the ace of spades and at least one more ace is: 3 48 3 48 3 48 | | A ∩ S = + + 1 3 2 2 3 1 Here the first term counts the number of hands with one additional ace, since there 3 1 48 are ways to choose the extra ace and ways to choose the other cards. Sim­ 3 ilarly, the second term counts hands with two additional aces, and the last term counts hands with all three remaining aces. In probability terms, we have: 2 3 2 3 1 48 48 3 3 48 3 + + 1 Pr (A ∩ S) = 52 5 Substituting these results into our original equation gives the solution: 2 3 2 51 3 1 48 48 3 3 48 3 + + 1 Pr (A S| ) 0.2214 . . . 4 (b) Suppose your friend says, “I have an ace.” What is the probability that she has another ace? Solution. The sample space and outcome probabilities are the same as before. Let L be the event that your friend has at least one ace, and M be the event that your friend has more than one ace. Our goal it to compute: Pr (M) Pr (M | L) = Pr (M ∩ L) = Pr (L) Pr (L) The second equality holds because your friend surely at least one ace if she has more than one; that is, M ⊆ L. The probability that your friend has at least one ace is: 4 1 48 + 4 2 48 4 3 48 + 4 4 48 + Pr (L) = 52 4 3 2 1 5