# ProbSetThree - Rensselaer Electrical, Computer, and Systems...

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Rensselaer Electrical, Computer, and Systems Engineering Department ECSE 4500 Probability for Engineering Applications PS#3 Solutions February 19, 2010 1. (1.35) Use Eq. (1.10-7) from the text with t =0 , t + τ =10 ,and λ ( τ ) as given. This gives Z 10 0 [1 e u/ 10 ] du e 1 =3 . 68 . Thus, P [ k clicks in 10 seconds ]= e 3 . 68 1 k ! (3 . 68) k . 2. (1.39) (a) Events associated with disjoint time intervals, under Poisson law, are independent. The number of cars arriving at a tollbooth in the time interval (0 ,T ) at a rate of λ per minute is such that the P [ k cars arrive in (0 )] = e λT [ λT ] k k ! . Let us de f ne the events: A , { n 1 cars arrive in (0 ,t 1 ) } , B , { n 2 cars arrive in ( t 1 ) } C , { n 1 + n 2 cars arrive in (0 ) } . We are asked to f nd the P [ A | C ] . From the de f nition of conditional probability, we know that this equals P [ AC ] P [ C ] . The event AC is the event that n 1 cars arrive in (0 1 ) and n 1 + n 2 cars arrive in (0 ) . This is the same as saying that n 1 cars arrive in (0 1 ) and n 2 cars arrive in ( t 1 ) , which is nothing but the event AB . Therefore, AC = AB .Bu t from the Poisson law (given), we know that P [ AB P [ A ] P [ B ] , because A and B are events on disjoint time intervals. Therefore, P [ A | C P [ AC ] P [ C ] = P [ AB ] P [ C ] = P [ A ] P [ B ] P [ C ] = ( λt 1 ) n 1 e λt 1 n 1 !

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## This note was uploaded on 04/15/2010 for the course ECSE 4500 taught by Professor Woods during the Spring '08 term at Rensselaer Polytechnic Institute.

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ProbSetThree - Rensselaer Electrical, Computer, and Systems...

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