# Probability of waiting for an interview

I have the following problem, for which I am having trouble finding a solution:

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 10:00, 10:15, and 10:30. Assume that the job candidates 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?

I think that the first applicant won't have to wait, the probability that the second has to wait is P(First Interview > 15min) and I am having trouble with the third.

My best guess is that he will have to wait if the sum of the remaining time for the first interviewee and the time for the second interviewee is greater than 15 minutes. Suppose that $X$ is first interviewee's remaining time, $Y$ is the second interviewee's time and $Z = X + Y$.

$$P(Z \ge 15) = 1 - P(Z \le 15)$$

and

$$P(Z \le 15) = \int_{0}^{15}\int_0^zf_x(x)f_y(z-x)dxdy$$

I'm not sure about that last integral. My idea is to using convolution integration to integrate everything that adds up to a certain sum and integrate that sum between 0 and 15. I'm having trouble with integration limits though. I used o and z in the inside integral because $z-x \ge 0$ (support of the exponential distribution) and so $z \ge x$.

• I suggest you just make $X$ denote the total length of the first interview, and then you want to calculate $P(Z\leq30)$. Jun 8 '14 at 23:12
• The way I interpret the problem, the 2nd interview cannot start until 10:15, even if the first interview is done before then. Hence, $Z = \text{max}(X,15)+Y$. Jun 8 '14 at 23:17
• The interviews don't start early, so total length won't work. I don't understand the $max(x, 15)$ part. The first interview's time up to 15 minutes plus the second interview's time? Jun 8 '14 at 23:33
• Sorry, I used a slightly different definition of $Z$. Define $Z$ to be the number of minutes after 10:00 that the 2nd interview ends. If $X \ge 15$, then $Z = X+Y$. If $X < 15$, then $Z = 15+Y$. This can be written more compactly as $Z = \text{max}(X,15)+Y$. Jun 8 '14 at 23:36

Let $X$ and $Y$ be the duration in minutes of the first and second interviews respectively. Then, the pdf of $X$ is $f_X(x) = \tfrac{1}{30}e^{-x/30}$, and the pdf of $Y$ is $f_Y(y) = \tfrac{1}{30}e^{-y/30}$. Assuming that the 2nd interview cannot start until the first interview is done AND the 2nd interviewee arrives at 10:15, the second interview ends at $Z = \text{max}(X,15)+Y$ minutes after 10:00. The probability of the 3rd interviewee waiting is the probability that $Z > 30$. Let's find $P(Z \le 30)$ instead.

If $Y > 15$, then $Z = \text{max}(X,15)+Y > 15+15 = 30$, so we must have $0 \le Y \le 15$.

If $X > 30-Y$, then $Z = \text{max}(X,15)+Y \ge (30-Y)+Y = 30$, so we must have $0 \le X \le 30-Y$.

Clearly, if $Y \le 15$ and $X \le 30-Y$, then $Z \le 30$.

Hence, the probability that $Z \le 30$ is given by:

$P(Z \le 30) = \dfrac{1}{30^2}\displaystyle\int_{0}^{15}\int_{0}^{30-y}e^{-x/30}e^{-y/30}\,dx\,dy \approx 0.20953$

Then, $P(Z > 30) = 1-P(Z \le 30) \approx 0.79047$.

• Thanks! I see what you're doing and I'm trying to reconcile it with what I'm doing. By the memorylessness of the distribution, the probability of the time remaining for $X$ at 10:15 is the same as if he just started. Therefore, at 10:15 it must be that $x+y \le 15$ and so $P(Z \le 15) = \int_{0}^{15}\int_0^{15-x}f_x(x)f_y(y)dydx$. However this gives me a different answer. How come? Jun 9 '14 at 6:36