A manager is interviewing 3 applicants for a job. The duration of each interview follows an exponential distribution with parameter 1/2, time being measured in hours.

The interviews are scheduled to begin at 8:00, 8:15, and 8:30. Assume that the job candidates (interviewees) arrive exactly on time.

For each of the three candidates, what is the probability that he/she will have to wait before his/her interview begins? (Provide 3 answers.)

How do I work out the answer for the third candidate?

  • $\begingroup$ Do you mean to say that the interviews are expected to last half an hour, but the interviewers only scheduled for fifteen minutes each? That's irresponsible planning. Back to your question: What do you know about the sum of two independent variables? $\endgroup$
    – Arthur
    Jun 9 '14 at 21:29
  • $\begingroup$ This was asked earlier today, I think. $\endgroup$ Jun 9 '14 at 21:42
  • $\begingroup$ Here. $\endgroup$ Jun 9 '14 at 21:43

There are two disjoint ways the third interview does not start at 8:30.

(i) The first interview lasts $15$ minutes or less and the second interview is longer than $15$ minutes or

(ii) The first interview lasts more than $15$ minutes, and the sum of the lengths is greater than $30$ minutes.

Calculating the probability of (i) is easy. Assume, unreasonably, that the first two interview lengths are independent. I take it that the parameter of the distribution is the mean. Let $X$ and $Y$ be the first two interview lengths. Then the probability of (i) is $\Pr(X\le 1/4)\Pr(Y\gt 1/4)$. Each of $\Pr(X\le 1/4)$ and $\Pr(Y\gt 1/4)$ is easy to calculate.

The probability of (ii) is harder. Recall that the exponential is memoryless*. So **given that $X\gt 1/4$, the additional time $X_1$ has exponential distribution with parameter $1/2$. Thus the probability of (ii) is $\Pr(X\ge 0.25)$ times the probability that $1/4+X_1+Y$ is greater than $\frac{1}{2}$.

So we need to find the probability that $X_1+Y$ is greater than $1/4$.

Note that $(X_1,Y)$ has joint density function $\frac{1}{4}e^{-x_1/2}e^{-y/2}$ when $x_1$ and $y$ are greater than $0$, and $0$ otherwise.

It is a little easier to calculate the probability $P$ that $X_1+Y\le 1/4$. For this, we need to integrate the joint density over the part of the plane that is below the line $x_1+y=1/4$. Thus $$P=\int_0^{1/4}\left(\int_{0}^{1/4-x_1} \frac{1}{4}e^{-x_1/2}e^{-y/2}\,dy \right)\,dx_1.$$ The probability of (ii) is then $\Pr(X\gt 1/4)(1-P)$.

Now add the probabilities of the two cases.

  • $\begingroup$ What if the first applicant is finished at 8:10, and the second applicant takes $17$ minutes to finish? $\endgroup$
    – Arthur
    Jun 9 '14 at 22:33
  • $\begingroup$ Thanks, my analysis was wrong, forgot about the fact people show up just on time. That complicates things, please see revised answer. $\endgroup$ Jun 9 '14 at 23:01

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