Much needed help.

A critical piece of surgical equipment has a failure rate of 86%. How many should we keep on hand in order to be 99.9% certain that we will have at least one piece which works?

I agree that it's very helpful to think of P(at least 1 works) as 1 - P(none of them works).

I just wanted to add that in order to use the formula for finding P(n failures) that JeffM is hinting at**, we need to make the assumption that the failures are independent, i.e. that if one of them fails it doesn't affect the odds of any of the others failing.
If the failures aren't independent then this would make this computation much more difficult (or impossible if the degree of dependence could not be ascertained).

This may seem like a trivial point at times, but it's really profoundly important to probability.

**I'm of course making an assumption here myself, which is that JeffM is referring to P(A₁, A₂, ... A_n) = P(A₁)xP(A₂)x...xP(A_n), where all A_k are independent.

JeffM said:

Not a trivial point at all. I have found, however, that to help a student work his or her own way through a problem, going one step at a time works best. Had the student followed the first step and responded, then would have been the time to discuss joint probability of independent events.

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Great point. Couldn't agree more. I'll definitely take that into consideration when responding to posts in the future.