There are n keys in the bundle, but only one fits the door's lock.

Trying them one after another, until the right one will be found, what is the probability it will happen on the k_{th} trial?

(In reply to

re: If the key fits ... (spoiler) by broll)

Well, yes, the conditional probability that the kth key will work, given that the first (k-1) keys did not, is indeed 1/(n-k+1). But I don't think that is what the problem is asking, especially as it is D1. Before we start trying keys, each key has an equal chance of working. I do agree that there is a small ambiguity in the question.

A more complicated way to reach my answer would be to calculate

the probability that the first k-1 keys did not work times the conditional probability that the kth key works given that the the first k-1 did not

= ((n-k+1)/n) * (1/(n-k+1)) = 1/n