# C three word messages arrive from transmitter a in a

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(c) Three-word messages arrive from transmitter A in a Poisson manner, with average rate λ A p W (3) = λ A / 6. Therefore, the random variable of interest is Erlang of order 8, and its PDF is given by f ( x ) = ( λ A / 6) 8 x 7 e λ A x/ 6 7! . (d) Every message originates from either transmitter A or B, and can be viewed as an independent Bernoulli trial. Each message has probability λ A / ( λ A + λ B ) of originating from transmitter A (view this as a “success”). Thus, the number of messages from transmitter A (out of the next twelve) is a binomial random variable, and the desired probability is equal to 12 8 λ A λ A + λ B 8 λ B λ A + λ B 4 . Solution to Problem 6.14. (a) Let X be the time until the first bulb failure. Let A (respectively, B ) be the event that the first bulb is of type A (respectively, B). Since the two bulb types are equally likely, the total expectation theorem yields E [ X ] = E [ X | A ] P ( A ) + E [ X | B ] P ( B ) = 1 · 1 2 + 1 3 · 1 2 = 2 3 . (b) Let D be the event of no bulb failures before time t . Using the total probability theorem, and the exponential distributions for bulbs of the two types, we obtain P ( D ) = P ( D | A ) P ( A ) + P ( D | B ) P ( B ) = 1 2 e t + 1 2 e 3 t . 72

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(c) We have P ( A | D ) = P ( A D ) P ( D ) = 1 2 e t 1 2 e t + 1 2 e 3 t = 1 1 + e 2 t . (d) We first find E [ X 2 ]. We use the fact that the second moment of an exponential random variable T with parameter λ is equal to E [ T 2 ] = E [ T ] 2 +var( T ) = 1 2 +1 2 = 2 2 . Conditioning on the two possible types of the first bulb, we obtain E [ X 2 ] = E [ X 2 | A ] P ( A ) + E [ X 2 | B ] P ( B ) = 2 · 1 2 + 2 9 · 1 2 = 10 9 . Finally, using the fact E [ X ] = 2 / 3 from part (a), var( X ) = E [ X 2 ] E [ X ] 2 = 10 9 2 2 3 2 = 2 3 . (e) This is the probability that out of the first 11 bulbs, exactly 3 were of type A and that the 12th bulb was of type A. It is equal to 11 3 1 2 12 . (f) This is the probability that out of the first 12 bulbs, exactly 4 were of type A, and is equal to 12 4 1 2 12 . (g) The PDF of the time between failures is ( e x + 3 e 3 x ) / 2, for x 0, and the associated transform is 1 2 1 1 s + 3 3 s . Since the times between successive failures are independent, the transform associated with the time until the 12th failure is given by 1 2 1 1 s + 3 3 s 12 . (h) Let Y be the total period of illumination provided by the first two type-B bulbs. This has an Erlang distribution of order 2, and its PDF is f Y ( y ) = 9 ye 3 y , y 0 . Let T be the period of illumination provided by the first type-A bulb. Its PDF is f T ( t ) = e t , t 0 . 73
We are interested in the event T < Y . We have P ( T < Y | Y = y ) = 1 e y , y 0 . Thus, P ( T < Y ) = 0 f Y ( y ) P ( T < Y | Y = y ) dy = 0 9 ye 3 y ( 1 e y ) dy = 7 16 , as can be verified by carrying out the integration. We now describe an alternative method for obtaining the answer. Let T A 1 be the period of illumination of the first type-A bulb. Let T B 1 and T B 2 be the period of illumination provided by the first and second type-B bulb, respectively. We are interested in the event { T A 1 < T B 1 + T B 2 } . We have P ( T A 1 < T B 1 + T B 2 ) = P ( T A 1 < T B 1 ) + P ( T A 1 T B 1 ) P ( T A 1 < T B 1 + T B 2 | T A 1 T B 1 ) = 1 1 + 3 + P ( T A 1 T B 1 ) P ( T A 1 T B 1 < T B 2 | T A 1 T B 1 ) = 1 4 + 3 4 P ( T A 1

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