Odds of outcome with P probability repeated N times 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?"
 A: 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 $8.22020473856651522786926141017182175041798 \times 10^{-6}$
A: Assume independence. This is not exactly a reasonable assumption, since a thunderstorm or worm may fry a lot of hard drives.
Under that assumption, the number $X$ of hard drive failures has binomial distribution, with $n=100$ and $p=0.01$.
The probability of exactly $k$ failures is
$$\binom{100}{k}p^k(1-p)^{100-k}.$$
For your problem, add up from $k=8$ to $k=100$.
If you are going to do this by hand, you can save considerable time by first finding the probability that $7$ or fewer hard drives fail. Call this number $f$. Then the probability of $8$ or more going bad is $1-f$. The expression for $f$ is
$$\small f=\binom{100}{0}(0.01)^0(0.99)^{100}+  \binom{100}{1}(0.01)^1(0.99)^{99}+ \cdots+ \binom{100}{7}(0.01)^7(0.99)^{93}.$$
Remarks:  With appropriate software, evaluation is easy. 
In the old days, this wasn't true, so good approximations were developed. In this case, the number of failures has a close to Poisson distribution, with $\lambda=(100)(0.01)$. The probability that a Poisson random variable with parameter $\lambda$ is equal to $k$ is $ e^{-\lambda}\dfrac{\lambda^k}{k!}$.
To get an excellent approximation to $f$, add up, $k=0$ to $k=7$. In our case, $\lambda=1$.  The first three terms will provide a good approximation. And one should not strive for great precision, since after all the nmber we started with $1\%$) is an estimate, and a probably quite unreliable one. For the percentage will change as your hard drives age. And the stress on hard drives may not be uniform from computer to computer. 
