Probability

Sheldon Ross's A First Course In Probability (Solution Manual) PDF

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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. e−λt + (1 − e−λt)p 71.

Condition on outcome of initial trial P(E before F) = P(E b FE)P(E) + P(E b FF)P(F) + P(E b Fneither E or F)[1 − P(E) − P(F)] = P(E) + P(E b F)(1 − P(E) − P(F)]. Hence, P(E b F) = 77. P( E ) . P( E ) + P( F ) (a) This is equal to the conditional probability that the first trial results in outcome 1 (F1) given that it results in either 1 or 2, giving the result 1/2. More formally, with L3 being the event that outcome 3 is the last to occur P(F1L3) = P( L3 F1 ) P ( F1 ) P ( L3 ) = (1/ 2)(1/ 3) = 1/ 2 1/ 3 (b) With S1 being the event that the second trial results in outcome 1, we have P(F1S1L3) = 78.

28 E[Y] = (40 + 33 + 25 + 50)/4 = 37 22. Let N denote the number of games played. (a) E(N) = 2[p2 + (1 − p)2] + 3[2p(1 − p)] = 2 + 2p(1 − p) The final equality could also have been obtained by using that N = 2 + ] where I is 0 if two games are played and 1 if three are played. Differentiation yields that d E[ N ] = 2 − 4 p dp and so the minimum occurs when 2 − 4p = 0 or p = 1/2. 48 Chapter 4 (b) E[N] = 3[p3 + (1 − p)3 + 4[3p2(1 − p)p + 3p(1 − p)2(1 − p)] + 5[6p2(1 − p)2 = 6p4 − 12p3 + 3p2 + 3p + 3 Differentiation yields d E[N ] = 24p3 − 36p2 + 6p + 3 dp Its value at p = 1/2 is easily seen to be 0.

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A First Course In Probability (Solution Manual) by Sheldon Ross


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