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Unformatted text preview: have 1Pr(XC>XB+XA)=1[Pr(XC>XB+XAXC>XA)Pr(XC>XA)+Pr(XC>XB+XA XC<XA)Pr(XC<XA)]=1Pr(XC>XB+XAXC>XA)Pr(XC>XA)Pr(XC>XB+XA XC<XA)Pr(XC<XA)=1Pr(XC>XB+XAXC>XA)Pr(XC>XA)0.Pr(XC<XA)=1Pr(XC>XB+XA XC>XA)Pr(XC>XA) Note that Pr(XC>XB+XAXC<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+XAXC>XA)=Pr(XC>XB) which simplifies PB to PB=1Pr(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.
 Winter '08
 Chisholm
 Probability

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