# Exponential distribution: Fuel pumps

There are 2 fuel pumps (one active and one reserve). If the first one fails, the reserve pump is brought online.

The life of fuel pumps follows an exponential distribution with an expected life of 2 days.

What is the probability that the system won't last for 1 full day?

I can solve this individually for each pump, but how do we do this for both pumps combined? I mean if first pump lasts 6 hours, the other would last less than 18 hours.. and so on.. How do we take care of this?

## 1 Answer

Let $X$ be the life length of the first pump, and let $Y$ be the life length of the second. Assume independence.

We want $\Pr(X+Y\lt 1)$.

The density function of $X$ is $\frac{1}{2}e^{-x/2}$ and the density function of $Y$ is $\frac{1}{2}e^{-y/2}$ (for positive $x$ and $y$). The joint density $f(x,y)$ of $X$ and $Y$ is therefore $\frac{1}{4}e^{-x/2}e^{-y/2}$ in the first quadrant, and $0$ elsewhere.

Draw the line $x+y=1$. We want to find the probability that $(X,Y)$ lands in the part of the first quadrant that is below this line.

That is a triangle $T$ with corners $(0,0)$, $(1,0)$, and $(0,1)$. Our required probability is $$\iint_T f(x,y)\,dx\,dy.$$ To evaluate, express as an iterated integral, say $$\int_{x=0}^1 \left(\int_{y=0}^{1-x} \frac{1}{4}e^{-x/2}e^{-y/2}\,dy\right)\,dx.$$ The integrations are mechanical.