Suppose that the time (in hours) required to repair a machine is an exponentially distributed random variable with parameter $\lambda = 0.2$. What is:

(a) the probability that a repair time exceeds 10 hours?
(b) the conditional probability that a repair takes at least 7 hours, given that it takes more than 6 hours?

I was able to answer part (a). The survival function of the exponential distribution is $e^{-\lambda t}$ so I plugged in for $\lambda = 0.2$ and $t = 10$ to calculate the probability that repair time is greater than $2$. I got $e^{-0.2\times 10} = .13533$ which is correct.

For part (b), I know I have to use the standard rules of conditional probability: $$ P(X > 7 | X > 6) = \frac{P(X > 7 \cap X > 6)}{P(X > 6)} $$ but for some reason my answer is incorrect.

Am I doing this incorrectly?


1 Answer 1


My guess is you have a cancellation error -- your approach seems correct.

The exponential random variable has the "memoryless property": given that we have seen no event up to time $t$, the conditional distribution of the remaining time until the event is the same as it originally was.

As you noted, if $X$ is the time until the first event, and we have times $t$ and $t + s$ (with $s \geq 0$), we have \begin{equation*}\begin{split} P(X > t + s | X > t) &= \frac{P(X > t + s)}{P(X > t)} \\ &= \frac{e^{-\lambda(t + s)}}{e^{-\lambda t}} \\ &= e^{-\lambda s} \\ &= P(X > s). \end{split}\end{equation*}


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