By Sheldon Ross

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

**Example text**

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.

6. n n P ∪ Ei = 1 − P ∩ Eic = 1 − 1 1 7. n ∏[1 − P( E )] i 1 (a) They will all be white if the last ball withdrawn from the urn (when all balls are withdrawn) is white. As it is equally likely to by any of the n + m balls the result follows. g g b P ( RBG G last) = . r +b+ g r +b+ g r +b bg b g Hence, the answer is . + (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).

Let Ei = {no type i in first n selections} N P{T > n} = P ∪ Ei i =1 = (1 − Pi ) n − ∑ ∑∑ (1 − P − P ) + ∑∑∑ (1 − p i j n I

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

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