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**Unformatted text preview: **Third Homework Set — Solutions Chapter 2 Problem 37 (a) There are ( 10 5 ) selections for the final exam. The number of selec- tions that allow the student to solve all problems is ( 7 5 ) , so that the desired probability is ( 7 5 ) ( 10 5 ) = 0 . 08333. (b) There are ( 7 4 ) · ( 3 1 ) selections that’ll let the student solve exactly four problems, so that the probability of solving at least four prob- lems is ( 7 5 ) + ( 7 4 ) · ( 3 1 ) ( 10 5 ) = 1 2 . Problem 43 (a) There are n ! ways to arrange n people in a line. There are 2( n- 1)! ways to arrange them such that A and B are next to each other. Hence, the probability of A and B being next to each other is 2( n- 1)! n ! = 2 n . (b) If n = 2, then A and B will always be next to each other. Now, assume that n > 3. After A picks a seat, there are n- 1 seats left, two of which are next to A , so that the desired probability is 2 n- 1 . Problem 50 The probability that you have five spades and your partner has the remaining eight spades is ( 13 5 )( 39 8 )( 8 8 )( 31 5 ) ( 52 13 , 13 , 26 ) = 2 . 6084 · 10- 6 ....

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