Question

Can someone please check my work? A man with 5 keys wants to open his door and tries the keys at random. Exactly one key will open the door. (a) What is the probability that he opened the door on his third trial if unsuccessful keys are eliminated? (b) What is the expected (mean) number of trials if unsuccessful keys are eliminated? MY ANSWERS: (a) Probability of having a (F)(F)(S) is (4/5)(3/4)(1/3) = .2 (b) E(X=1)+E(X=2)+E(X=3)+E(X=4)+E(X=5) (1/.2)+(1/.25)+(1/(1/3))+(1/.5)+(1/1) = 15 I feel like my part b is unreasonable. I used the mean formula for the binomial distribution (1/p)