Poisson Process Probability of a machine failing before other

Question

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?

Answer 1

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}