Question

A man has a bunch of n keys, only one of which exactly open the lock. The man tries to open the lock of using the key randomly. The probability that he open the lock at the ${\text{j}}^{th}$ attempt by assuming that rejected key has already tried is

• A. $\frac{\text{j}}{n}$
• B. $\frac{1}{n}$
• C. ${(1-\frac{1}{n})}^{j-1}\times \frac{1}{n}$
• D. None of these

Answer 1

The correct option is B $\frac{1}{n}$

Probability that lock open is first trial is $1$

Probability that lock open in second trail is $1.\frac{n-1}{n}$

Probability that lock open in third trial is $1.\frac{n-1}{n}.\frac{n-2}{n-1}$

Required probability $=\frac{n-1}{n}.\frac{n-2}{n-1}\cdots \frac{n-j+1}{n-j+2}.\frac{1}{n-j+1}=\frac{1}{n}$