# Poisson Process Probability of a machine failing before other

Machine $1$ is working now. Machine $2$ will be switched on at time $t$. Suppose that machine $1$ fails at rate $λ_1$ and $2$ at rate $λ_2$with an exponential waiting time. What is the probability that machine $2$ fails first?

I thought:

It should be $P(X_2+t<X_1)$ or $P(X_2<X_1|X_2=t)$ but I don't know how to go about calculating it.

Can someone give me a hint?

• Calculate the probability M1 fails before time $t$. Once both are working, probability M1<M2 is $\frac {\lambda_1}{\lambda_1+\lambda_2}$ – A.S. Mar 4 '16 at 10:17
• This is definitely $P(X_2+t<X_1)$ (I cannot even fathom what $P(X_2<X_1|X_2=t)$ stands for...) and one knows that, for every nonnegative $x$, $P(X_1>x)=e^{-\lambda_1x}$ hence, by independence, $P(X_1>X_2+t)=E(e^{-\lambda_1(X_2+t)})=e^{-\lambda_1t}E(e^{-\lambda_1X_2})$. Now, can you compute $E(e^{-sX_2})$ for every $s$? – Did Mar 4 '16 at 10:17
• @A.S. would this be simply $P(X_1>t)P(X_1>X_2)$? – GRS Mar 4 '16 at 10:25
• Yes. I was computing the complementary probability instead. Your answer is coincides with Did's term-for-term. – A.S. Mar 4 '16 at 10:27
• @Did I had a think about it, but I don't understand why we are taking expectation? I know how to calculate $E(X_2)=1/\lambda_2$ – GRS Mar 4 '16 at 10:27

Let $X_i$ be the failure time of machine $i$. By lack of memory we have \begin{align} \mathbb P(X_2+t<X_1) &= \mathbb P(X_1>X_2+t\mid X_1>t)\mathbb P(X_1>t)\\ &= \mathbb P(X_1>X_2)\mathbb P(X_1>t)\\ &= \left(\frac{\lambda_2}{\lambda_1+\lambda_2}\right) e^{-\lambda_1 t}. \end{align}