Question

a). 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 the probability the door opens with the third key he

tries is also(1/n )

.

b). Let(n=50). In R, calculate the conditional probability of opening the door with each of

the keys(k=2, . . . 50) given that the previous keys did not work. Additionally, calculate the

probability of opening the door with exactly the second, third, . . . , 50th key.

Answer 1