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By Sheldon Ross

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I + 1)! n  n + 1 1 (n + 1) p ∑  i + 1  p 1 ( n + 1) p ∑  i +1 (1 − p) n − i i =0 n +1  n + 1 j n +1− j  p (1 − p ) j  j =1   n + 1 0 1 n +1− 0  1 −  0  p (1 − p)  (n + 1) p     1 = [1 − (1 − p ) n +1 ] (n + 1) p = 11. For any given arrangement of k successes and n − k failures: P{arrangementtotal of k successes} = 12. 1 P{arrangement} p k (1 − p ) n − k = = n  P{k successes}  n  k n−k   p (1 − p)   k  k  Condition on the number of functioning components and then use the results of Example 4c of Chapter 1: n Prob = n ∑  i  p (1 − p) i =0 i n − i  i + 1   n   n − i   i       i +1 where   = 0 if n − i > i + 1.

Assume n > 1. 2 (a) n (b) Conditioning on whether the man of couple j sits next to the woman of couple i gives the 1 1 n−2 2 2n − 3 result: + = n − 1 n − 1 n − 1 n − 1 (n − 1) 2 (c) e−2 68. exp(−10e−5} 69. 2212 54 Chapter 4 70. e−λt + (1 − e−λt)p 71.  26  (a)    38  5 3  26  12 (b)    38  38 72. 4) 3   73. Let N be the number of games played. 8125 74. 2 (a)   3 5 5 3 6 2 7 8  8  2   1   8  2   1   8  2   1   2  (b)      +      +      +    5  3   3   6  3   3   7  3   3   3  5  5  2  1 (c)     4  3  3  6  2  (d)     4  3  5 1   3 2 76.

P{X = 0} = P{1 loses to 2} = 1/2 P{X = 1} = P{of 1, 2, 3: 3 has largest, then 1, then 2} = (1/3)(1/2) = 1/6 P{X = 2} = P{of 1, 2, 3, 4: 4 has largest and 1 has next largest} = (1/4)(1/3) = 1/12 P{X = 3} = P{of 1, 2, 3, 4, 5: 5 has largest then 1} = (1/5)(1/4) = 1/20 P{X = 4} = P{1 has largest} = 1/5 Chapter 4 47 15. P{X = 1} = 11/66  12 − j  11  66  54 + j =2 11 P{X = 2} = ∑  P{X = 3} = ∑∑  k ≠1 k≠ j P{X = 4} = 1 −   j   11  12 − j  12 − k     66  54 + j  42 + j + k  j =2 3 ∑ P{X = 1} i =1 16.

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