A dime is tossed repeatedly until a head appears. Let N be the trial number on which this first head occurs. Then a nickel is tossed N times. Let X count the number of times that the nickel comes up tails. Determine $Pr(X = x)$ and $E[X]$.
I can see that N is a geometric distribution with p=1/2. So,
$$P(N=n)=\frac{1}{2^n}$$ with $n \ge 1$ (or should it be $n \ge 0$?)
$$P(X=x)=\sum^\infty_{n=1}{n \choose x}\frac{1}{2^{2n}}$$
Or should $n$ start from 0? Although, with 0, the sum would be greater than 1.
Then, should $E[X]$ be:
$$E[X] = \sum^\infty_{x=0}x\sum^\infty_{n=x+1}{n \choose x}\frac{1}{2^{2n}}$$
Or should it be $n=x$?
So as you can see, I'm having trouble with the boundary of $n$. Should it start from 1 or 0?
There is a 50% chance that there will be 1 trial (the dime lands on heads)
There is a 25% chance that there will be 2 trials (the dime lands on tails-heads)
There is a 12.5% chance that there will be 3 trials (the dime lands on tails-tails-heads)
There is a 6.25% chance that there will be 4 trials (the dime lands on tails-tails-tails-heads)
...
Does this help?