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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 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
1nk+1
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B
1nk1
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C
1n
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D
1n1
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Solution

The correct option is B 1n
Total number of keys =n
Probability of success in first trial is 1n.
So, probability of failure in first trial is 11n=n1n.

Now, if he failed in unlocking with first key, he tries with the second key.
So, total number of keys left =n1

Probability of unlocking in 2nd trial is

=P(he fails in first trial)×P(he succeeds in second trial)
=n1n1n1

Probability of failure in 2nd trial is n2n1.

So, the probability that the person unlocks the lock at kth trial. It means that he fails in first (k1) trials.

=P(he fails in k-1 trials)×P(he succeeds in k^{th} trial)

=[n1n.n2n1.n3n2....n(k1)n(k2)]×[1n(k1)]

=1n

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