By Sheldon Ross

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

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

**Example text**

1 − (1 − p)n ≥ 1/2, or, n ≥ − Chapter 3 log(2) log(1 − p ) 41 i −1 12. ai ∏ (1 − a j ) is the probability that the first head appears on the ith flip and ∞ ∏ (1 − a ) is the i i =1 j =1 probability that all flips land on tails. 13. Condition on the initial flip. If it lands on heads then A will win with probability Pn−1,m whereas if it lands tails then B will win with probability Pm,n (and so A will win with probability 1 − Pm,n). 14. Let N go to infinity in Example 4j. 15. P{r successes before m failures} = P{rth success occurs before trial m + r} m + r −1 n − 1 r n−r = p (1 − p ) .

1 1 1 1 3 4+2+3 9 (1) + + 3 32 34 45. P{5thheads} = P{heads 5th }P{5th } ∑ P{h i th }P{i th } i 5 1 1 = 1010 10 = . i 1 11 i =1 10 10 ∑ 46. 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.

R − 1 n=r ∑ 16. If the first trial is a success, then the remaining n − 1 must result in an odd number of successes, whereas if it is a failure, then the remaining n − 1 must result in an even number of successes. 17. P1 = 1/3 P2 = (1/3)(4/5) + (2/3)(1/5) = 2/5 P3 = (1/3)(4/5)(6/7) + (2/3)(4/5)(1/7) + (1/3)(1/5)(1/7) = 3/7 P4 = 4/9 (b) Pn = n 2n + 1 (c) Condition on the result of trial n to obtain Pn = (1 − Pn−1) 1 2n + Pn −1 2n + 1 2n + 1 (d) Must show that n n −1 1 n − 1 2n = 1− + 2n + 1 2n − 1 2n + 1 2n − 1 2n + 1 or equivalently, that n n 1 n − 1 2n = + 2n + 1 2n − 1 2n + 1 2n − 1 2n + 1 But the right hand side is equal to n + 2n(n − 1) n = (2n − 1)(2n + 1) 2n + 1 42 Chapter 3 18.

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

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