Question

6. X and Y are lifetimes of two computers in an office. X and Y are independent and identically distributed exponential variables with mean lifetimes of 5 years. What is the probability that both computers will become inoperable within 4 years

Answer 1

Both random variables have the same distribution with density function

$f_X(x) = \begin{cases}\frac15e^{-x/5} & \text{if } x \ge 0 \\ 0 & \text{otherwise}\end{cases}$

and since X and Y are independent, their joint density is the product of their marginal densities,

$f_{X,Y}(x,y) = \begin{cases}\frac1{25}e^{-(x+y)/5} & \text{if } x\ge0 \text{ and } y \ge 0 \\ 0 & \text{otherwise}\end{cases}$

Then the probability that both computers become inoperable within 4 years is

$\mathrm{Pr}[X<4\text{ and }Y<4] = \displaystyle \int_{-\infty}^4 \int_{-\infty}^4 f_{X,Y}(x,y) \, dx \, dy$

$\mathrm{Pr}[X<4\text{ and }Y<4] = \displaystyle \frac1{25} \int_0^4 \int_0^4 e^{-(x+y)/5} \, dx \, dy$

$\mathrm{Pr}[X<4\text{ and }Y<4] = \displaystyle \frac1{25} \left(\int_0^4 e^{-x/5}\,dx\right)\left(\int_0^4 e^{-y/5} \, dy\right)$

$\mathrm{Pr}[X<4\text{ and }Y<4] = \displaystyle \frac1{25} \left(\int_0^4 e^{-x/5}\,dx\right)^2$

$\mathrm{Pr}[X<4\text{ and }Y<4] = \displaystyle \frac1{25} \left(5-\frac5{e^{4/5}}\right)^2 \approx \boxed{0.3032}$