My professor did the following calculation in class:
- Suppose you concatenate $n$ hard drives of size $m$ with annualized probability of failure $x$ into one superdrive of size $m \cdot n$
The probability of at least one drive in the superdrive failing is
$$1 - (1 - x)^n$$
I understand this. $1 - x$ is the probability of one not failing, $(1-x)^n$ is the probability of all not failing (assuming independence). We subtract it from $1$ to get the probability of at least one failing.
Now suppose that the the super drive does not fail unless at least two drives fail, and that there is a Mean Time To Repair of 1 day.
He stated that the probability of failure is then
$$(x \cdot n) \cdot (x \cdot (n-1)) \frac{1}{365}$$
I do not understand this calculation. Could someone please break it down?
Specifically, why does $(x \cdot n)$ give us the probability any one drive failures, and $(x \cdot (n-1))$ give us the probability that any second drive fails. I understand that $n$ is the number of drives that can fail and $n - 1$ is the number of remaining drives that can fail, but why can we simply multiply by the probability of a single drive failing $x$? Additionally, why can we simply divide by $\frac{1}{365}$? Does this assume uniformity?