# Time for first disk failure which is exponentially distributed

This is not homework, but exam preparation.

Suppose disk $A$ fails in an exponential distribution with parameter $\lambda$ (meaning, in average it fails every $\frac{1}{\lambda}$ time), and disk $B$ fails in an exponential distribution with parameter $\mu$. Disks $A$ and $B$ are independent.

I am asked what is the average time it takes for the first failure to happen. I tackled the problem in the following way:

The probability both disks fail up until time $t$ is $P(A\leq t)P(B\leq t)$, which has distribution of $(1-e^{-\lambda t})(1-e^{-\mu t})$. The probability both disks survive past time $t$ is $P(A \geq t)P(B \geq t) = (1-P(A\leq t))(1-P(B\leq t))=e^{-\lambda t}e^{-\mu t}$. So, the probability of the first failure is distributed $1 - (1-e^{-\lambda t})(1-e^{-\mu t}) - e^{-\lambda t}e^{-\mu t}$, and the average time it takes to happen is just the expected value calculation of that distribution

However, I see that a student before me tried to solve it this way, and the grader said that this is completely wrong with no elaboration.

What is wrong in this way?

You are supposed to find the mean of the distribution of $\min(A,B)$.
$$P(\min(A,B) > t) = P(A>t)P(B > t) = (1-P(A \leq t))(1-P(B \leq t)) = e^{-\lambda t}e^{-\mu t}$$
$$f_{\min(A,B)}(t) = \frac{d P(\min(A,B)\leq t)}{dt} = (\lambda+\mu)e^{-(\lambda+\mu)t}$$
$$E(\min(A,B)) = \frac{1}{\lambda+\mu}$$