Probability Theory; Binonial Distribution?

**The problem statement**

Now you and your fiend play a different game. You flip your coin until it comes up heads the first time. Let X denote the number of flips needed. Your friend rolls its die until it comes up "3" or "5". The first try let Y denote the number of rolls needed. Assume X and Y are independent; note that X >= 1 and Y >=1. a) Determine P(X=n), n>=1 b) Determine P(Y=n), n>=1 c) Determine E(x) d) Determine E(y) e) Determine P(x=y) f) Given that x = y, determine the expected value of this common value.

**Equations**

Binomial Distribution

nCk (P)^k (1-P)^n-k

**The attempt at a solution**

a) nC1 (1/2)(1/2)^n-1 (My friend did not include the nC1)

b) nC1 (1/3)(2/3)^n-1 (My friend did not include the nC1)

c) n*1/2

d) n*1/3

e) nC1 (1/2)(1/2)^n-1 * nC1 (1/3)(2/3)^n-1 (This is weird, my friend integrated from 1 to infinite) he got 1/4??

f) nsubx*psubx + nsuby*psuby (I definitely need help on this one.)

Update:

Can someone confirm these?

a) (1/2)(1/2)^n-1

b) (1/3)(2/3)^n-1

c) E(X)=1/(1/2)=2

d) E(Y)=1/(2/6)=3

e) P(x=y)= sum(n=1 to +oo) 1/6(1/3)^(n-1) = 1/4

f) sum(n=1 to +oo) n*(1/6)*(1/3)^(n-1) = 3/8