Q.45 - A First Course in Probability

Question

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

(a) If she tries the keys at random, discarding those that do not work, what is the probability that she will open the door on her kth try? 

(b) What if she does not discard previously tried keys? 

Step-by-Step Solution

Answer

a). The probability she will open the door on her $k$th try is $\frac{1}{n}$.

b). The probability she does not discard previously tried keys is $\frac{1}{n}{\left(1-\frac{1}{n}\right)}^{k-1}$.

Part (a) Step 1: Given Information

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

Part (a) Step 2: Explanation

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

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

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 $n!$

$P\left(\text{ correct key is }i\text{-th }\right)=\frac{(n-1)!}{n!}=\frac{1}{n}$

Part (b) Step 1: Given Information

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

Part (b) Step 2: Explanation

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 $n^k$

$P(\text{ correct key is }i-th)=\frac{(n-1{)}^{k-1}}{n^k}=\frac{1}{n}{\left(1-\frac{1}{n}\right)}^{k-1}$