combined reliability I thought I knew the answer to this question, but further reflection is showing some holes in my knowledge; my college math is twenty-five years old and google isn't helping today. Let's say you have a computer with five disk drives. If any one of the drives crashes, the computer is down. For this model of drives, 99% of them run a year without crashing. So the probability that this computer will be up the entire year is > .99 number of drives > > or > > .99 5 ≈ .95 That's correct, right? Now the obverse observation is that each drive has a 1% chance of **failure** over the course of a year. How does one express the total chance of failure for the whole computer? The answer should be 5%, because that's what's left over from the 95% given above. The naive approach would be to do the same thing again: > .015 but that gives 0.0000000001 which obviously isn't right. What am I missing here?

$0.01^5$ is the probability that ALL 5 drives crash.

To see why, consider that an individual drive failing is unaffected by what happens with the other drives, and the probability of any one of them failing is 0.01. So the probability of all of them failing is the product of each probability.

To calculate the probability of failure, you do exactly as you have done, calculate 1 - Probability of not failing.