T b 1 t b 2 t a 1 t b 1 given the event t a 1 t b 1

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T B 1 < T B 2 | T A 1 T B 1 ) . Given the event T A 1 T B 1 , and using the memorylessness property of the exponential random variable T A 1 , the remaining time T A 1 T B 1 until the failure of the type-A bulb is exponentially distributed, so that P ( T A 1 T B 1 < T B 2 | T A 1 T B 1 ) = P ( T A 1 < T B 2 ) = P ( T A 1 < T B 1 ) = 1 4 . Therefore, P ( T A 1 < T B 1 + T B 2 ) = 1 4 + 3 4 · 1 4 = 7 16 . (i) Let V be the total period of illumination provided by type-B bulbs while the process is in operation. Let N be the number of light bulbs, out of the first 12, that are of type B . Let X i be the period of illumination from the i th type-B bulb. We then have V = Y 1 + · · · + Y N . Note that N is a binomial random variable, with parameters n = 12 and p = 1 / 2, so that E [ N ] = 6 , var( N ) = 12 · 1 2 · 1 2 = 3 . Furthermore, E [ X i ] = 1 / 3 and var( X i ) = 1 / 9. Using the formulas for the mean and variance of the sum of a random number of random variables, we obtain E [ V ] = E [ N ] E [ X i ] = 2 , and var( V ) = var( X i ) E [ N ] + E [ X i ] 2 var( N ) = 1 9 · 6 + 1 9 · 3 = 1 . 74
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(j) Using the notation in parts (a)-(c), and the result of part (c), we have E [ T | D ] = t + E [ T t | D A ] P ( A | D ) + E [ T t | D B ] P ( B | D ) = t + 1 · 1 1 + e 2 t + 1 3 1 1 1 + e 2 t = t + 1 3 + 2 3 · 1 1 + e 2 t . Solution to Problem 6.15. (a) The total arrival process corresponds to the merging of two independent Poisson processes, and is therefore Poisson with rate λ = λ A + λ B = 7. Thus, the number N of jobs that arrive in a given three-minute interval is a Poisson random variable, with E [ N ] = 3 λ = 21, var( N ) = 21, and PMF p N ( n ) = (21) n e 21 n ! , n = 0 , 1 , 2 , . . . . (b) Each of these 10 jobs has probability λ A / ( λ A + λ B ) = 3 / 7 of being of type A, inde- pendently of the others. Thus, the binomial PMF applies and the desired probability is equal to 10 3 3 7 3 4 7 7 . (c) Each future arrival is of type A with probability λ A / ( λ A + λ B ) = 3 / 7, independently of other arrivals. Thus, the number K of arrivals until the first type A arrival is geometric with parameter 3 / 7. The number of type B arrivals before the first type A arrival is equal to K 1, and its PMF is similar to a geometric, except that it is shifted by one unit to the left. In particular, p K ( k ) = 3 7 4 7 k , k = 0 , 1 , 2 , . . . . (d) The fact that at time 0 there were two type A jobs in the system simply states that there were exactly two type A arrivals between time 1 and time 0. Let X and Y be the arrival times of these two jobs. Consider splitting the interval [ 1 , 0] into many time slots of length δ . Since each time instant is equally likely to contain an arrival and since the arrival times are independent, it follows that X and Y are independent uniform random variables. We are interested in the PDF of Z = max { X, Y } . We first find the CDF of Z . We have, for z [ 1 , 0], P ( Z z ) = P ( X z and Y z ) = (1 + z ) 2 . By differentiating, we obtain f Z ( z ) = 2(1 + z ) , 1 z 0 . (e) Let T be the arrival time of this type B job. We can express T in the form T = K + X , where K is a nonnegative integer and X lies in [0,1]. We claim that X 75
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is independent from K and that X is uniformly distributed. Indeed, conditioned on the event K = k , we know that there was a single arrival in the interval [ k, k + 1].
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