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) f

Answer 1

@Data_LG2

Answer 2

i forgot how to do probabilities :/

maybe @mathmate knows.

Answer 3

I agree with your part (a)
(b) expected value is \(\sum xP(x)\)
Here x is the number of tries (from one to 5) and P(x) is the probability of choosing the right key after x tries (same way you calculated part (a).
Sum x over 1 to 5, and the result is E[X].

Answer 4

@BrighterDays

Answer 5

i.e., in your terminology:
E[X]=1*P(X=1)+2*P(X=2)+3*P(x=3)+4*P(X=4)+5*P(X=5)
The sum of the 5 probabilities \(\sum P(X=i)\) must equal exactly 1.

Answer 6

Great! Thank you!!

Answer 7

You're welcome!

Answer 8

@mathmate - I have another one I don't know where to start with. I'll post it and tag you. If you wouldn't mind looking at it, I'd appreciate it!