EXAMPLE 5.8

Suppose that the lifetimes of two components are independent of one another and that the first lifetime, $X_1$, has an exponential distribution with parameter ${\lambda }_1$, whereas the second, $X_2$, has an exponential distribution with parameter ${\lambda }_2$. Then the joint pdf is

$$\begin{array}{rl}& f\left(x_1,x_2\right)\\ & =f_{X_1}\left(x_1\right)\cdot f_{X_2}\left(x_2\right)\\ & =\{\begin{array}{ll}{\lambda }_1{e}^{-{\lambda }_1{x}_1}\cdot {\lambda }_2{e}^{-{\lambda }_2{x}_2}={\lambda }_1{\lambda }_2{e}^{-{\lambda }_1{x}_1-{\lambda }_2{x}_2}& \text{ if }{x}_1>0,x_2>0\\ 0& \text{ otherwise }\end{array}\end{array}$$

Let ${\lambda }_1=1/1000$ and ${\lambda }_2=1/1200$, so that the expected lifetimes are 1000 hours and 1200 hours, respectively. The probability that both component lifetimes are at least 1500 hours is

$$\begin{array}{rl}& P\left(1500\le X_1,1500\le X_2\right)\\ & =P\left(1500\le X_1\right)\cdot P\left(1500\le X_2\right)\\ & =e^{-{\lambda }_1\left(1500\right)}\cdot e^{-{\lambda }_2\left(1500\right)}\\ & =\left(0.2231\right)\left(0.2865\right)=\mathrm{0.0639.}\end{array}$$