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Two widget repairmen work in the same shop. Repair order completion times for the first worker are exponentially distributed, with parameter 3 per hour. Completion time for the second worker are exponentially distributed with parameter 5 per hour. A repair order was submitted 20 minutes ago and has yet completed. Find the probability that the order was assigned to the first worker.

This is a practice problem for an exam I am studying so the exact answer is not as important as a proper solution.

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up vote 1 down vote accepted

By the way, the second worker is more efficient, so there is some incompatibility between the title and the text. We will use the text, and look for the conditional probability that the assignment was given to the first worker.

We need more information! Suppose as an extreme case that the second worker, though fast, gets very few repair assignments. That will strongly affect the calculation. We are probably invited to assume that assignments are given to the two workers with probability $1/2$. This assumption is probably unreasonable.

Let $A$ be the event the assignment was given to the first worker, and let $N$ be the event "not completed." We want $\Pr(A|N)$. Use the standard formula $$P(A|N)=\frac{\Pr(A\cap N)}{\Pr(N)}.$$ We want to compute the two probabilities on the right.

We first find the "harder" probability $\Pr(N)$. The event $N$ can happen in $2$ ways: (i) assignment was given to first worker, and is not completed or (ii) assignment was given to second worker, and is not completed.

To find the probability of (i), the probability an assignment is given to the first worker is, by assumption, $1/2$. Given that this happened, the probability of non-completion is the probability that a certain exponential is $\gt 1/3$ (hour).

Use a similar calculation for (ii). Add (i) and (ii) to find $\Pr(N)$.

Finally, note that $\Pr(A\cap N)$ is ust the already computed probability of (i).

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