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}`