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Since the second is known to be white, there are only 3 ways out of the remaining 8 in which the first can be white, so that the probability is 3 > 8.
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1.43. The probabilities that a husband and wife will be alive 20 years from now are given by 0.8 and 0.9, respectively. Find the probability that in 20 years (a) both, (b) neither, (c) at least one, will be alive.
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Let H, W be the events that the husband and wife, respectively, will be alive in 20 years. Then P(H) 0.8, P(W) 0.9. We suppose that H and W are independent events, which may or may not be reasonable. (a) P(both will be alive) (b) P(neither will be alive) P(H > W ) 1 P(H)P(W ) (0.8)(0.9) 1 0.72. 0.02. 0.98. 0.02 P(Hr > Wr) P(Hr) P(Wr) (0.2)(0.1)
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(c) P(at least one will be alive)
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P(neither will be alive)
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1.44. An inefficient secretary places n different letters into n differently addressed envelopes at random. Find the probability that at least one of the letters will arrive at the proper destination.
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Let A1, A2, . . . An denote the events that the 1st, 2nd, . . . , nth letter is in the correct envelope. Then the event that at least one letter is in the correct envelope is A1 < A2 < c < An, and we want to find P(A1 < A2 < c < An). From a generalization of the results (10) and (11), page 6, we have (1) P(A1 < A2 < c < An) a P(Ak) c a P(Aj > Ak ) a P(Ai > Aj > Ak)
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( 1)n 1P(A1 > A2 > c > An)
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where a P(Ak ) the sum of the probabilities of Ak from 1 to n, a P(Aj > Ak) is the sum of the probabilities of Aj > Ak with j and k from 1 to n and k j, etc. We have, for example, the following: (2) P(A1) 1 n and similarly P(Ak) 1 n
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since, of the n envelopes, only 1 will have the proper address. Also (3) P(A1 > A2) P(A1) P(A2 u A1) 1 anb a 1 n 1 b
since, if the 1st letter is in the proper envelope, then only 1 of the remaining n similar way we find (4) P(A1 > A2 > A3) P(A1) P(A2 u A1) P(A3 u A1 > A2) 1 anb a n
1 envelopes will be proper. In a ba b
1 n 2
CHAPTER 1 Basic Probability
etc., and finally (5) P(A1 > A2 > c > An) 1 anb a 1 n 1 1 b c a b 1 1 n!
n Now in the sum a P(Aj > Ak) there are a b nC2 terms all having the value given by (3). Similarly in 2 n a P(Ai > Aj > Ak), there are a b nC3 terms all having the value given by (4). Therefore, the required 3 probability is P(A1 < A2 < c < An) n 1 a b anb 1 c 1 1 2! n 1 1 a b anb a b n 1 2 n 1 ( 1)n 1 a b a b n n! 1 3! c ( 1)n
n 1 1 1 a b anb a ba b n 1 n 2 3
1 n!
From calculus we know that (see Appendix A) ex so that for x 1 e or
x2 2!
x3 3!
1 1 2!
a1 1 3!
1 2! c
1 3! 1
cb e
It follows that if n is large, the required probability is very nearly 1 e 1 0.6321. This means that there is a good chance of at least 1 letter arriving at the proper destination. The result is remarkable in that the probability remains practically constant for all n 10. Therefore, the probability that at least 1 letter will arrive at its proper destination is practically the same whether n is 10 or 10,000.
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