8 - For an alternative approach, note that the event of...

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For an alternative approach, note that the event of interest occurs if and only if the time Y 2 of the second arrival is less than or equal to 2. Hence, the desired probability is P ( Y 2 2) = Z 2 0 f Y 2 ( y ) dy = Z 2 0 (0 . 6) 2 ye 0 . 6 y dy. This integral can be evaluated by integrating by parts, but this is more tedious than the ±rst approach. (d) The expected number of ±sh caught is equal to the expected number of ±sh caught during the ±rst two hours (which is 2 λ =2 · 0 . 6=1 . 2), plus the expectation of the number N of ±sh caught after the ±rst two hours. We have N = 0 if he stops ±shing at two hours, and N = 1, if he continues beyond the two hours. The event { N =1 } occurs if and only if no ±sh are caught in the ±rst two hours, so that E [ N ]= P ( N = 1) = P (0 , 2) = 0 . 301. Thus, the expected number of ±sh caught is 1 . 2+0 . 301 = 1 . 501 . (e) Given that he has been ±shing for 4 hours, the future ±shing time is the time until the ±rst ±sh is caught. By the memorylessness property of the Poisson process, the future time is exponential, with mean 1 . Hence, the expected total ±shing time is 4+(1 / 0 . 6) = 5 . 667. Solution to Problem 6.11. We note that the process of departures of customers who have bought a book is obtained by splitting the Poisson process of customer departures, and is itself a Poisson process, with rate . (a) This is the time until the ±rst customer departure in the split Poisson process. It is therefore exponentially distributed with parameter . (b) This is the probability of no customers in the split Poisson process during an hour, and using the result of part (a), equals e . (c) This is the expected number of customers in the split Poisson process during an hour, and is equal to . Solution to Problem 6.12. Let X be the number of di²erent types of pizza ordered. Let X i be the random variable de±ned by X i = n 1 , if a type i pizza is ordered by at least one customer, 0 , otherwise. We have X = X 1 + ··· + X n ,and E [ X n E [ X 1 ]. We can think of the customers arriving as a Poisson process, and with each customer independently choosing whether to order a type 1 pizza (this happens with probability 1 /n ) or not. This is the situation encountered in splitting of Poisson pro- cesses, and the number of type 1 pizza orders, denoted Y 1 , is a Poisson random variable with parameter λ/n .W ehav e E [ X 1 P ( Y 1 > 0) = 1 P ( Y 1 =0)=1 e λ/n , so that E [ X n E [ X 1 n ( 1 e λ/n ) . 71
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Solution to Problem 6.13. (a) Let R be the total number of messages received during an interval of duration t . Note that R is a Poisson random variable with arrival rate λ A + λ B . Therefore, the probability that exactly nine messages are received is P ( R =9)= ( ( λ A + λ B ) T ) 9 e ( λ A + λ B ) t 9! . (b) Let R be deFned as in part (a), and let W i be the number of words in the i th message. Then, N = W 1 + W 2 + ··· + W R , which is a sum of a random number of random variables. Thus, E [ N ]= E [ W ] E [ R ] = ± 1 · 2 6 +2 · 3 6 +3 · 1 6 ² ( λ A + λ B ) t = 11 6 ( λ A + λ B ) t.
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8 - For an alternative approach, note that the event of...

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