Question

A man has n keys on a key ring, one of which opens the door to his apartment. Having celebrated a bit too much one evening, he returns home only to find himself unable to distinguish one key from another. Resourcesful, he works out a fiendishly clever plan: He will choose a key at random and try it. If it fails to open the door, he will discard it and choose at random one of the remaining n-1 keys, and so on. Clearly, the probability that he gains entrance with the first key he selects is 1/n. What is the probability that the door opens with the third key

Answer 1

The value is
`P(3) = (1)/(n)`

Explanation:

From the question we are told that

The number of keys is n

The number of keys remaining after the first key is chosen is n-1

The probability that he gains entrance with the first key he selects is
`(1)/(n)`

Generally the probability that the first key does not open the door is

`p(F_1) = 1 - (1)/(n) = (n-1)/(n)`

Generally the number of keys remaining after the second key is chosen is

n-2

Generally the probability that he gains entrance with the second key he selects is

`(1)/(n-1)`

Generally the probability that the second key does not opens the door is

`P(F_2) = 1- (1)/(n-1) = (n-2)/(n-1)`

Generally the probability that he gains entrance with the third key he selects is

`(1)/(n-2)`

Generally the probability that the door opens with the third key

`P(3) = p(F_1) * P(F_2) * (1)/(n-2)`

=>
`P(3) =  (n-1)/(n) *   (n-2)/(n-1) * (n-2)/(n-1)`

=>
`P(3) =  (1)/(n)`

Note :

All the outcome of the events are independent