[math] \newcommand{\NA}{{\rm NA}} \newcommand{\mat}[1]{{\bf#1}} \newcommand{\exref}[1]{\ref{##1}} \newcommand{\secstoprocess}{\all} \newcommand{\NA}{{\rm NA}} \newcommand{\mathds}{\mathbb}[/math]

Jones puts in two new lightbulbs: a 60 watt bulb and a

100 watt bulb. It is claimed that the lifetime of the 60 watt bulb has an exponential density with average lifetime 200 hours ([math]\lambda = 1/200[/math]). The 100 watt bulb also has an exponential density but with average lifetime of only 100 hours ([math]\lambda = 1/100[/math]). Jones wonders what is the probability that the 100 watt bulb will outlast the 60 watt bulb.

If [math]X[/math] and [math]Y[/math] are two independent random variables with exponential densities [math]f(x) = \lambda e^{-\lambda x}[/math] and [math]g(x) = \mu e^{-\mu x}[/math], respectively, then the probability that [math]X[/math] is less than [math]Y[/math] is given by </math> P(X < Y) = \int_0^\infty f(x)(1 - G(x))\,dx,

[[math]] where $G(x)$ is the cumulative distribution function for \ltmath\gtg(x)[[/math]]

. Explain why this is the case. Use this to show that </math> P(X < Y) = \frac \lambda{\lambda + \mu} <math>$ and to answer Jones's question.