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?

  • 1
    $\begingroup$ Calculate the probability M1 fails before time $t$. Once both are working, probability M1<M2 is $\frac {\lambda_1}{\lambda_1+\lambda_2}$ $\endgroup$ – A.S. Mar 4 '16 at 10:17
  • $\begingroup$ 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 $s$? $\endgroup$ – Did Mar 4 '16 at 10:17
  • $\begingroup$ @A.S. would this be simply $P(X_1>t)P(X_1>X_2)$? $\endgroup$ – GRS Mar 4 '16 at 10:25
  • $\begingroup$ Yes. I was computing the complementary probability instead. Your answer is coincides with Did's term-for-term. $\endgroup$ – A.S. Mar 4 '16 at 10:27
  • $\begingroup$ @Did I had a think about it, but I don't understand why we are taking expectation? I know how to calculate $E(X_2)=1/\lambda_2$ $\endgroup$ – 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}


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.