If I have 100 hard drives, and the odds of a hard drive failing on a given week are 1%, what is the probability of 8 (or more) hard drives failing on the same week?
I realize this is a fairly basic probability question but I can't quite figure out how to approach the problem and which combinations/permutations/binomials/products etc to use to "Frame" the problem mathematically.
The best I can come up with is a more generalized statement of the problem: "Given an event with probability P, what is the probability Q of the event occurring at least N times out of T trials?"
Best Answer
For this, you would want to use multiple binomial distributions; for this, you would have to sum N elements, each of the following form -
$\binom{n}{k} \times (P)^k \times (1 - P)^{n - k}$
Where k is the number of trials - in this case, you would sum this little formula with k as eight, then k as nine, then k as ten, all the way to 100. A mathematical translation of this is as follows -
$\sum_{i = 8}^{100} \binom{100}{i} \times (0.01)^i \times (1 - 0.01)^{100 - i}$
The reason this works is that, for each iteration, you are finding the chance that a set amount of hardwares do not fail (which is the second term in the product), the chance that a set amount of hardwares fail (which is the third term), and then the number of combinations of hardwares this is possible with (the first term).
A shorter way to calculate may be to sum from zero to seven, and then subtract that from one, as P(Less than 8 hardwares failing) = 1 - P(At least 8 hardwares failing) - this would be written as follows -
$1 - \sum_{i = 1}^{7} \binom{100}{i} \times (0.01)^i \times (1 - 0.01)^{100 - i}$
Either way, the answer will be about $8.22020473856651522786926141017182175041798 \times 10^{-6}$