Probability of multiple events Suppose I have doorman who answers the door 10% of the time when someone knocks. I'm trying to derive a formula that gives the probability on the nth knock. For getting the chance of continuous success, it seems trivial. But how do I compute it in a way that's "the chance one of the first 5 knocks will be answered"? I thought maybe just n/10, but that seems too silly easy. This is also likely very trivial, so I apologize.

It's one minus the probability that the door doesn't get answered on the first five knocks. The probability that it doesn't get answered on any one knock is $9/10$ so your answer is

$$1-\Bigl( \frac{9}{10}\Bigr)^5.$$

This assumes that whether or not the doorman opens the door on the first knock is independent of whether he opens it on the second knock, etcetera.