91Ó°ÊÓ

A woman has $n$ keys, of which one will open her door.

Trying keys and discarding them ( $n$ keys):

Outcome space: $S=\left\{x_1,x_2,…x_n;\text{some permutation of keys}\right\}$

Each element of the outcome space is equally likely.

The probability that the key is $i$-th in the permutation is the number of permutations with fixed $i$-th place, and that is $(n-1)!$ divided by the number of permutations $n!$

A woman has $n$ keys, of which one will open her door.

b) Trying keys and returning them them ( $n$keys)

Each element of the outcome space is equally likely.

The probability that the key is $i$-th in the vector is the number of vectors where the first $n-1$ elements are some of the $n-1$ keys that are not correct and the last key is the correct one, and that is $(n-1{)}^{k-1}$ divided by the number of elements in the outcome space $n^k$