A, B and C student arrive at the beginning of a professor's office time. The duration of time they will stay is exponentially distributed with means of 1, 1/2,and 1/3 hour.
I want to find the probability they are the last student left for each student and the expected time until all three students are gone.
I tried to transform it into a problem of Markov chain. We may denote the state of the Markov chain by the rates of the students that are left and use $\emptyset$ denote an empty office. We get the following matrix:
$\begin{pmatrix}
-6 & 3 & 2 & 1 \\
0 & -3 & 0 & 0 \\
0 & 0 & -4 & 0 \\
0 & 0 & 0 & -5
\end{pmatrix}$
How would one continue to do this problem? Are there more efficient way to do this?
Best Answer
There is no Markov chain here, since there is no change from one state to another one.
Formally, the problem is the following: take $X, Y, Z$ independent, with $X \sim \mathcal{E}(1)$, $Y \sim \mathcal{E}(1/2)$, $X \sim \mathcal{E}(1/3)$. Then
The probability Student A leaves first is $\mathbb{P}(X=\min(X,Y,Z))$. This is known for exponential distributions.
The expected time that all three students are gone is $\mathbb{E}(\max(X,Y,Z))$. Can you figure out how to compute this?