Question

Exactly one of six similar keys opens a certain door. If you try the keys, one after another, what is the expected number of keys that you will have to try before success?

Answer 1

P(Success) = 1/6

If he succeed in first trial probability is 1/6

If he succeed in secod trial probability is 5/6 * 1/5 (Where 5/6 is probability of failure)

If he succeed in third trial, probability is 5/6 * 4/5 * 3/4

and so on ...

E(X) = 1 * 1/6 + 2 * (5/6 * 1/5) + 3 * (5/6 * 4/5 * 1/4) + 4 * (5/6 * 4/5 * 3/4 * 1/3)

+ 5 * (5/6 * 4/5 * 3/4 * 2/3 * 1/2) + 6 * (5/6 * 4/5 * 3/4 * 2/3 * 1/2 * 1)

= 1/6 + 2 * 1/6 + 3 * 1/6 + 4 * 1/6 + 5 * 1/6 + 6 * 1/6

= 7/2

Expected number of keys = 7/2