Find $P(Y=1)=?$

Question

Let $X$ be the number of tosses of a fair coin required to get the first head. If $Y | X = n$ is distributed as $Binomial(n,$1/2$)$, then what is $P(Y = 1)$ ?

Answer 1

The following calculations are rough because you have not specified which geometric distribution you are using. $$P(Y=1) = P(Y=1|X=x)P(X=x) = \sum_x {x \choose 1} (1/2)^x P(X=x)$$ $$=\sum_xx(1/4)^x = \frac{1/4}{(3/4)^2} = 4/9$$

The rough idea is to apply Bayes and the appropriate formula for whichever geometric distribution you are using. Then you will get an infinite sum which should be straightforward to get in closed form.