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Unformatted text preview: Solutions to Homework 7 Section 6.1 8. In order to solve this problem, we need to compute the number of poker hands that contain the ace of hearts. There is no choice about choosing the ace of hearts. To form the rest of the hand, we need to choose 4 cards from the 51 remaining cards, so there are C (51 , 4) hands containing the ace of hearts. Therefore the answer to the question is the ratio C (51 , 4) C (52 , 5) = 5 52 . 16. Of the C (52 , 5) = 2 , 598 , 960 hands, 4 C (13 , 5) = 5148 are flushes, since we can specify a flush by choosing a suit and then choosing 5 cards from that suit. Therefore the answer is 5148 / 2598960 = 33 / 16660 . 0020 . 24. In each case, if the numbers are chosen from the integers from 1 to n, then there are C ( n, 6) possible entries, only one of which is the winning one, so the answer is 1 /C ( n, 6) . (a) 1 /C (30 , 6) 1 . 7 10 6 (d) 1 /C (48 , 6) 8 . 1 10 8 34. (a) There are 50 49 48 47 equally likely outcomes of the drawings. In only one of these do Bo, Colleen, Jeff, and Rohini win the first, second, third, and fourth prizes, respectively. Therefore the probability is 1 / (50 49 48 47) = 1 / 5527200 . (b) There are 50 50 50 50 equally likely outcomes of the drawings. In only one of these do Bo, Colleen, Jeff, and Rohini win the first, second, third, and fourth prizes, respectively. Therefore the probability is 1 / 50 4 = 1 / 6250000 . Section 6.2 1 8. (a) Since 1 has either to precede 2 or to follow it, and there is no reason that one of these should be any more likely than the other, we immediately see that the answer is 1 / 2 . (c) For 1 immediately to precede 2, we can think of these two numbers as glued together in forming the permutation. Then we are really permuting n 1 numbers the single numbers from 3 through n and the one glued object, 12. There are ( n 1)! ways to do this. Since there are n ! permutations in all, the probability of randomly selecting one of these is ( n 1)! /n ! = 1 /n . 18. We assume that births are independent and the probability of a birth in each day is 1 / 7 . (a) The probability that the second person has the same birth dayoftheweek as the first person is 1 / 7 . (b) The probability that all the birth daysoftheweek are different if p n = 6 7 5 7 8 n 7 since each person after the first must have a different birth dayofthe week from all the previous people in the group. Note that if n 8 , then p n = 0 since the seventh fraction is 0. The probability that at least two are born on the same day of the week is therefore 1 p n ....
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This note was uploaded on 06/07/2011 for the course COT 3100 taught by Professor Staff during the Summer '08 term at University of Florida.
 Summer '08
 Staff

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