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. Resourceful, 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. Show that theprobability the door opens with the third key he tries is also 1/n.

Answer 1

Answer and Step-by-step explanation:

The probability that he gains entrance with the first key he selects is 1/n. Show that theprobability the door opens with the third key he tries is also 1/n.

get in 1st try

right key = 1

keys in the chain: n

p = 1/n

get in 3rd try means wrong key 1st try and wrong key 2nd try:

wrong keys 1st try = n - 1

keys not tried in the chain = n

wrong keys 2nd try = n - 2

keys not tried in the chain = n - 1

right key 3rd try = 1

keys not tried in the chain = n - 2

1st 2nd 3rd

n-1/n * n-2/n-1 * 1/n-2 = 1/n