Q20-Ch5 - have

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Regarding a question asked in the tutorial, the following is some clarification about part b of question 20 in chapter 5 we discussed in the tutorial class today. To make the notations simple, let XC=the time that person C spends in server 1. Similarly XA and XB are the times that A and B spend in server 2, respectively. We know from the question that XC~exp(lambda1) and XA, XB ~exp(lambda2). Also, all of the random variables are independent of each other. part.b) PB=Pr(B is still in the system when C is done with server1)=1-Pr(both A and B have already left before C is done with server1)=1-Pr(XC>XB+XA) now to solve this, let's condition on the presence of A in the system when C is done with server1. We
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Unformatted text preview: have 1-Pr(XC>XB+XA)=1-[Pr(XC>XB+XA|XC>XA)Pr(XC>XA)+Pr(XC>XB+XA| XC<XA)Pr(XC<XA)]=1-Pr(XC>XB+XA|XC>XA)Pr(XC>XA)-Pr(XC>XB+XA| XC<XA)Pr(XC<XA)=1-Pr(XC>XB+XA|XC>XA)Pr(XC>XA)-0.Pr(XC<XA)=1-Pr(XC>XB+XA| XC>XA)Pr(XC>XA) Note that Pr(XC>XB+XA|XC<XA)=0 since XB is positive (exponential random variable). In the last term, since all the random variables are exponentially distributed, we apply the memoryless property of the exponential distribution soPr(XC>XB+XA|XC>XA)=Pr(XC>XB) which simplifies PB to PB=1-Pr(XC>XB)Pr(XC>XA) The rest was done in the tutorial class....
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This note was uploaded on 03/17/2012 for the course STAT 333 taught by Professor Chisholm during the Winter '08 term at Waterloo.

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