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By Sheldon Ross

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

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53 64. (a) 1 − 7 ∑e −4 i 4 / i! ≡ p i =0 (b) 1 − (1 − p)12 − 12p(1 − p)11 (c) (1 − p)i−1p 65. (a) 1 − e−1/2 1 1 − e −1 / 2 − e −1 / 2 2 (b) P{X ≥ 2X ≥ 1} = 1 − e −1 / 2 (c) 1 − e−1/2 (d) 1 − exp {−500 − i)/1000} 66. Assume n > 1. 2 (a) 2n − 1 2 (b) 2n − 2 (c) exp{−2n/(2n − 1)} ≈ e−1 67. 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.

With C denoting this event, and Oi the event that the ith offspring is albino, we have: P(O1) = P(O1C)P(C) + P(O1Cc)P(Cc) = (1/4)(2/3) + 0(1/3) = 1/6 (b) P(O2 O1c ) = = P(O1cO2 ) P(O1c ) P(O1cO2 C ) P(C ) + P(O1cO2 C c ) P(C c ) 5/6 (3 / 4)(1 / 4)(2 / 3) + 0(1 / 3) 3 = = 5/6 20 62. (a) P{both hitat least one hit} = P{both hit} P{at least one hit} = p1p2/(1 − q1q2) (b) P{Barb hitat least one hit} = p1/(1 − q1q2) Qi = 1 − pi, and we have assumed that the outcomes of the shots are independent.

Also, let A be the event that she is accepted and R that she is rejected. 4) 25 53. Let W and F be the events that component 1 works and that the system functions. P(WF) = 55. P{Boy, F} = P(WF ) P(W ) 1/ 2 = = c P( F ) 1 − P( F ) 1 − (1 / 2) n −1 4 16 + x P{Boy) = so independence ⇒ 4 = 10 16 + x P{F} = 10 16 + x 10 ⋅ 10 ⇒ 4x = 36 or x= 9. 16 + x A direct check now shows that 9 sophomore girls (which the above shows is necessary) is also sufficient for independence of sex and class. 56. P{new} = ∑ P{new i 30 type i}pi = ∑ (1 − p ) i n −1 pi i Chapter 3 57.

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