Minimum of Two Exponential Random Variables

Question

Elevator $A$ breaks down with an exponential distribution of $4$ per day, while elevator $B$ breaks down with an exponential distribution of $6$ per day. Assume a day begins at midnight. What is the probability that the first breakdown occurs before $2$ AM?

My attempt:

Let $X$ = the time between two consecutive breakdowns for elevator A and $Y$ = the time between two consecutive breakdowns for elevator B. Then $\min(X,Y)$ is exponentially distributed with rate $10$. Thus, let $Z = \min(X,Y)$, so $P(Z < 2) = 1-e^{-20} = 1$ approximately. My only issue with this is that $Z$ is the minimum time between two consecutive breakdowns, not a single breakdown.. Aren't I overcompensating? For example, my answer includes the probability that the elevator breaks down at $5$ AM and then breaks down again at $6$ AM. Also, the probability is $1$, which seems wrong as well.

Answer 1

The number of breakdowns of the first elevator in a day has a Poisson distribution with a mean of $4.$ The thing that has an exponential distribution is the time until the next breakdown, which has an expected value of $1/4\text{ day}.$ With the two elevators together the mean waiting time is $1/10\text{ day}.$

Since $2\text{ hours} = 1/12\text{ day},$ the probability that it happens within that time is $1- e^{-(1/12)/(1/10)} = 1 - e^{-10/12} \approx 0.5654.$

One would speak here not of the minimum of two exponential distributions, but of the minimum of two exponentially distributed random variables.

Answer 2

Note that the rates are $4$ breakdowns per $24$ hours and $6$ breakdowns per $24$ hours. Therefore, $\lambda_1 = \frac{4}{24}$, and $\lambda_2 = \frac{6}{24}$. Using them, you obtain $$\Pr\{Z\le 2\} = 1-\exp(-20/24).$$