Question

There is a key-ring which has $n$ keys of which only one is the right key of the lock. A person tries to open the lock at random. If he discards the key already tried, what is the probability that he opens the lock at $k^{th}$ trial?

• A. $\frac{1}{n-k+1}$
• B. $\frac{1}{n-k-1}$
• C. $\frac{1}{n}$
• D. $\frac{1}{n-1}$

Answer 1

Answer: (C)

$\frac{1}{n}$

Total number of keys $=n$

Probability of success in first trial is $\frac{1}{n}$.

So, probability of failure in first trial is $1-\frac{1}{n}=\frac{n-1}{n}$.

Now, if he failed in unlocking with first key, he tries with the second key.

So, total number of keys left $=n-1$

Probability of unlocking in 2nd trial is

$=P\left(\text{he fails in first trial}\right)\times P\left(\text{he succeeds in second trial}\right)$

$=\frac{n-1}{n}\frac{1}{n-1}$

Probability of failure in 2nd trial is $\frac{n-2}{n-1}$.

So, the probability that the person unlocks the lock at $k^{th}$ trial. It means that he fails in first ($k-1$) trials.

$=P\left(\text{he fails in k-1 trials}\right)\times P\left(\text{he succeeds in k^{th} trial}\right)$

$=[\frac{n-1}{n}.\frac{n-2}{n-1}.\frac{n-3}{n-2}....\frac{n-(k-1)}{n-(k-2)}]\times \left[\frac{1}{n-(k-1)}\right]$

$=\frac{1}{n}$