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Below find the solution C 6 min=0.1 hour the time in hours from the start of the interval until the firstlog-on. X has an exponential distribution with u03bb = 25 P( X > 0.1) = u222b 25e^(u2212 25 x) dx = e ^(u2212 25( 0.1)) = 0.082 P(X=0)=?In 6 minutesX: Number of log-ons in 6 minutesE(X)=(0.4167(log-ons)/(min))(6min)rarr E(X)=2.5 log-ons=lambdaP(X=x)=(e^(-lambda)lambda^(x))/(x!)x=0rarr P(X=0)=(e^(-2.5)(2.5)^(0))/(0!)=((0.082)(1))/(1)rarrP(X=0)=0.082B) P ... See the full answer