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27. P{X = n + kX > n} = P{ X = n + k} P{ X > n} p(1 − p) n + k −1 (1 − p ) n = p(1 − p)k−1 = If the first n trials are fall failures, then it is as if we are beginning anew at that time. 28. 29. The events {X > n} and {Y < r} are both equivalent to the event that there are fewer than r successes in the first n trials; hence, they are the same event. P{ X = k + 1} P{ X = k}  Np  N − np     k + 1 n − k − 1  =  Np  N − Np      k  n − k  = Chapter 4 ( Np − k )(n − k ) (k + 1)( N − Np − n + k + 1) 61 30.

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 ) . 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.

2) 29 42. P{Afailure} = 43. 1 (1) 4 4 3 P{2 headedheads} = = = . 1 1 1 1 3 4+2+3 9 (1) + + 3 32 34 45. P{5thheads} = 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(A2A1) = = = 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.

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