I flip a fair coin. What is the expected number of attempts it will take for one heads to show up and one tails to show up?
Attempt at a solution: I have been trying to use conditional probability, with E(X) = (sigma from k = 1 to infinity)(E(X|Y = k))(P(Y = k))
Any help is appreciated.
Ok. So please read up
Coupon collector's problem - Wikipedia, the free encyclopedia
Answer to your question is a direct application of that formula.
Another way to look at it is
There are to mutually exclusive and exhaustive events
E1: H....... (i.e. first flip is H)
E2: T....... (i.e first flip is T)
p(E1)=p(E2)=0.5
Consider only E1. After the 1st flip you need a T. What is the expected number of flips to get a T? Let this be n. (What is n?)
So expected number of flips to get atleast H and T = n+1
Similarly for E2. (use symmetry)
Final answer = p(E1)(n+1) + p(E2)(n+1) = n+1.