The probability distribution of waiting time until two exponentially distributed events with different parameters both occur

Question

I am working on a problem related to the waiting time until a parking garage is empty. We are given that the cars independently spend an exponential distributed time in the parking garage, with parameter $\mu$. We are at a point where the garage is closed and no more cars are arriving, but two cars remain. The problem is to find the probability distribution of the waiting time until the garage is empty. Conveniently, we are given the solution. Letting $T$ denote the waiting time, we have $$ f_T(t) = 2 \mu e^{-\mu t} \left(1 - e^{-\mu t}\right). $$ However, I do not understand how to arrive at this answer.

Here is my work so far: Let $X$ denote the number of cars in the parking garage. We may model this as a death process, where the death rate at $X = 2$ is $2\mu$, and then at $X=1$ it is $\mu$. Then the waiting time for the first car to leave is exponentially distributed with parameter $2\mu$, i.e. $$ f_{\text{Car 1}}(t) = 2\mu e^{-2\mu t}, $$ and similarily $$ f_{\text{Car 2}}(t) = \mu e^{-\mu t}. $$ I believe it is correct so far. However, I am not sure on how to proceed from here. Any help or hints are greatly appreciated.

Answer 1

Let $T_1$ and $T_2$ be the waiting times for each car. You are trying to find $T \equiv \max (T_1, T_2)$. For this type of problem it is fairly simple to obtain the result by working with the CDFs. Assuming that the waiting times for each car are independent you can use the following rule:

$$\begin{equation} \begin{aligned} F_T(t) \equiv \mathbb{P}(T \leqslant t) &= \mathbb{P}(T_1 \leqslant t, T_2 \leqslant t) \\[6pt] &= F_{T_1}(t) F_{T_2}(t) \\[6pt] &= (1-\exp(-\mu t)) (1-\exp(-\mu t)) \\[6pt] &= 1-2\exp(-\mu t)+\exp(-2\mu t). \\[6pt] \end{aligned} \end{equation}$$

(Note that this is just an application of the rule for failure times in a parallel system in reliability analysis.) Hence, you have:

$$\begin{equation} \begin{aligned} f_t(t) = \frac{d F_T}{dt}(t) &= 2 \mu \exp(-\mu t) - 2 \mu \exp(-2\mu t) \\[6pt] &= 2 \mu \exp(-\mu t) (1 - \exp(-\mu t)). \\[6pt] \end{aligned} \end{equation}$$