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

By Sheldon Ross

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**Example text**

R + b)(r + b + g ) r + b + g r + g (b) P(RBG) = 8. (a) P(A) = P(AC)P(C) + P(AC c)P(C c) > P(BC)P(C) + P(BC c )P(C c) = P(B) (b) For the events given in the hint P(C A) P ( A) (1/ 6)(1/ 6) = = 1/ 3 P(AC) = 3/ 36 3/ 36 Because 1/6 = P(A is a weighted average of P(AC) and P(ACc), it follows from the result P(AC) > P(A) that P(AC c) < P(A). Similarly, 1/3 = P(BC) > P(B) > P(BC c) However, P(ABC) = 0 < P(ABC c). 9. P(A) = P(B) = P(C) = 1/2, P(AB) = P(AC) = P(BC) = 1/4. But, P(ABC) = 1/4.

54. 2 Chapter 4 55. 2 e + e 2 2 56. The number of people in a random collection of size n that have the same birthday as yourself is approximately Poisson distributed with mean n/365. Hence, the probability that at least one person has the same birthday as you is approximately 1 − e−n/365. Now, e−x = 1/2 when x = log(2). Thus, 1 − e−n/365 ≥ 1/2 when n/365 ≥ log(2). That is, there must be at least 365 log(2) people. 57. (a) 1 − e−3 − 3e−3 − e−3 32 17 = 1 − e−3 2 2 17 1 − e−3 P{ X ≥ 3} 2 = (b) P{X ≥ 3X ≥ 1} = P{ X ≥ 1} 1 − e−3 59.

Let M and F denote, respectively, the events that the policyholder is male and that the policyholder is female. Conditioning on which is the case gives the following. P(A2A1) = = = P( A1 A2 ) P( A1 ) P( A1 A2 M )α + P( A1 A2 F )(1 − α ) P( A1 M )α + P( A1 F )(1 − α ) pm2 α + p 2f (1 − α ) pmα + p f (1 − α ) Hence, we need to show that pm2 α + p 2f [1 − α ) > (pmα + pf(1 − α))2 or equivalently, that pm2 (α − α 2 ) + p 2f [1 − α − (1 − a) 2 ] > 2α(1 − α)pfpm Chapter 3 27 Factoring out α(1 − α) gives the equivalent condition pm2 + p 2f > 2 pf m or (pm − pf)2 > 0 which follows because pm ≠ pf.