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55PS7 - SOLUTION SET 7DUE Please report any errors in this...

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SOLUTION SET 7—DUE 3/19/2008 Please report any errors in this document to Ian Sammis ( [email protected] ). Problem 1 (#6.1.14) . What is the probability that a five-card poker hand contains cards of five different kinds? Solution. The sample space here is the space of poker hands—unordered collections of 5 of the 52 cards of the deck. There are ( 52 5 ) such hands, each with probability p = 1 52 5 « . There are ( 13 5 ) ways to choose 5 different kinds to make up a poker hand. Each card can be of any of the four suits, and since the cards are all distinguishable (they’re of different kinds), there are 4 5 ways to assign suits to the cards, for a total of ( 13 5 ) 4 5 such hands. Thus, the probability of getting a poker hand with 5 different kinds of cards is p (5 different kinds) = ( 13 5 ) 4 5 52 5 Problem 2 (#6.1.28) . In a superlottery, a player selects 7 numbers out of the first 80 positive integers. What is the probability that a person winds the grand prize by picking 7 numbers that are among the 11 numbers selected at random by a computer? Solution. There are several ways of solving this problem, depending upon which sample space you decide to use. Let’s consider as a sample space the ( 80 11 ) ways that the computer can pick winning numbers, each with probability 1 80 11 « . For the player to win, the computer must choose the player’s 7 numbers, along with any other 4 from amongst the remaining 73. There are ( 73 4 ) ways to do so, so the odds of winning the superlottery are p (win) = 73 4 80 11 = 73! 69!4! 80! 69!11! = 73!11! 80!4! . If you chose a different sample space (say, the ( 80 7 ) ways that the player can play, or the ( 80 7 ) × ( 80 11 ) ordered pairs of the form (player choices, computer choices), your work may look very different, but your final answer should be exactly the same. Problem 3 (#6.1.36) . Which is more likely: rolling a total of 8 when two dice are rolled or rolling a total of 8 when three dice are rolled? Solution. When two dice are rolled, the sample space consists of ordered pairs ( d 1 , d 2 ) of the faces of each die. There are 36 such pairs, 5 of which—(2,6), (3,5), (4,4), (5,3), and (6,2)—sum to 8. The odds of getting an 8 on two dice is thus 5 36 0 . 1389. 1

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2 SOLUTION SET 7—DUE 3/19/2008 When three dice are rolled, the sample space consists of ordered triplets ( d 1 , d 2 , d 3 ). There are 6 3 = 216 such triplets. We’re interested in the case d 1 + d 2 + d 3 = 8. Since the d i 1, we can replace them with new variables δ i = d i - 1, resulting in the equation δ 1 + δ 2 + δ 3 = 5. Each of the d i must be no greater than 6, but fortunately this is enforced for us, since the sum of the delta is itself 5. Thus we can now just make a standard stars and bars argument, and see that 8 can be rolled ( 5+3 - 1 5 ) = ( 7 5 ) = 21 ways. Thus, the probability of rolling an 8 on three dice is 21 216 0 . 09722. Thus it’s more likely to roll an 8 on two dice than on three.
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