By Sheldon Ross

**Read or Download A First Course In Probability (Solution Manual) PDF**

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**Extra resources for A First Course In Probability (Solution Manual)**

**Sample text**

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{arrangementtotal 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.