Question

a thief has a bunch of n keys, exactly one of which fits a lock. if the thief tries to open the lock by trying the keys at random, what is the probability that he requires exactly k attempts, if he rejects the keys already tried?

Answer 1

The probability that a thief requires exactly k attempts to open a lock with n keys, trying at random without retrying keys, involves finding the product of (n-1)/n, (n-2)/(n-1), ..., 1/(n-k+1). This is a geometric probability problem involving combinatorial logic.

Step-by-step explanation:

The question is asking for the probability that a thief requires exactly k attempts to open a lock with one correct key out of n keys, under the condition that any tried keys are not reused. This is a classic problem in probability, relying on the understanding of geometric distributions and combinatorial reasoning.

If the correct key is tried on the kth attempt, it means that the previous k-1 attempts have been failures (trying the wrong key) and the kth attempt is a success.

To solve this, consider that there are (n-1) incorrect keys among the n total keys. The probability that the first key tried is incorrect is (n-1)/n, the second key tried is incorrect (after one wrong key has already been removed) is (n-2)/(n-1), and this continues until the (k-1)th attempt.

When it comes to the successful attempt, the probability that the correct key is tried is 1/(n-k+1). Multiplying these probabilities together gives us the overall probability of the correct key being tried on exactly the kth attempt:

1. (n-1)/n (failure on the 1st attempt)
2. (n-2)/(n-1) (failure on the 2nd attempt)
3. ...
4. 1/(n-k+1) (success on the kth attempt)

The resulting probability is given by the product of these fractions, which represents the probability that the thief requires exactly k attempts to find the correct key, with all previous attempts being unsuccessful.