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Question: There is a key-ring which has n keys of which only one is the right key of the lock. A person tries ...

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 kth trial?
A. 1n−k+1\dfrac{1}{{n - k + 1}}
B. 1n−k−1\dfrac{1}{{n - k - 1}}
C. 1n\dfrac{1}{n}
D. 1n−1\dfrac{1}{{n - 1}}

Explanation

Solution

Hint: We start solving this question by using formula P = favourable case/total number of cases. We have to find the probability of the kth trial, so to find the required probability, we will find the probability for first trial, then second trial. Taking these cases will help us in finding a series so we can find the probability of kth trial.

Complete step-by-step answer:
Now, it is given that there is only one key out of n keys which can open the lock. We have to find the probability if the person opens the lock in the kth trial. We will start taking different cases to solve this problem.
Assuming if the person opens the lock in the first trial, then probability = favourable cases/total cases.
Therefore, probability of success in the first trial = 1n\dfrac{1}{n}
Also, probability of failure in the first trial = 1 - 1n\dfrac{1}{n} = n−1n\dfrac{{n - 1}}{n}
Now, if the person fails with the first key, he will try with the second key. Also, according to the question, the key already tried is discarded. So, the number of keys left = n – 1.
So, probability of success in second trial = P (fails in first trial) x P (success in second trial)
Therefore, probability of success in second trial = n−1n×1n−1\dfrac{{n - 1}}{n} \times \dfrac{1}{{n - 1}}
Also, probability of failure in the second trial = 1 - 1n−1\dfrac{1}{{n - 1}} = n−2n−1\dfrac{{n - 2}}{{n - 1}}
Now, taking the kth trial. For kth trail, keys left = n – k – 1
So, the probability of success in the kth trial = [n−1n.n−2n−1.....n−(k−1)n−(k−2)]×1n−(k−1)\left[ {\dfrac{{n - 1}}{n}.\dfrac{{n - 2}}{{n - 1}}.....\dfrac{{n - (k - 1)}}{{n - (k - 2)}}} \right] \times \dfrac{1}{{n - (k - 1)}}
So, probability that he opens the lock in the kth trail = 1n\dfrac{1}{n}
So, option (C) is correct.

Note: Whenever we come up with such problems, where we have to find the probability, we will start by taking different cases and then find the probability with the cases. Solving such problems makes the question easy and it helps in getting the right answer. Cases help us in seeing the patterns, just like in this question, we can see that there is a pattern of probability at every trial.