Probability question? 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?
 A: 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. 
