Let $p$ be the probability that the tip points downward after a person throws a drawing pin once. Miriam throws a drawing pin until it points downward for the first time. Let $X$ be the number of throws for this to happen. She then throws the drawing pin another $X$ times. Let $Y$ be the number of times the drawing pin points downward in the latter series of throws. Find the distribution of $Y$.
$Y$ should be binomial $(x,p)$ and $P(Y=0)=(1-p)^x$. Is there anything missing how $P(Y=0)=1-P/2-P$?
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Given $X=x$, the conditional distribution of $Y$ is $\operatorname{Bin}(x,p)$. $X$ itself follows $\operatorname{Geo}(p)$.
$$\begin{aligned} P(Y=0) & =\sum_{x=1}^{\infty} P(Y=0|X=x)P(X=x)\\ & =\sum_{x=1}^{\infty} (1-p)^{x}(1-p)^{x-1}p \\ & = p(1-p)^{-1}\sum_{x=1}^{\infty} (1-p)^{2x}\\ & =p(1-p)^{-1}\frac{(1-p)^2}{1-(1-p)^2}=\frac{1-p}{2-p} \end{aligned}$$
You are correct that given $X=x$, you have $Y$ with a binomial distribution with parameters $x$ and $p$
The distribution for $X$ is geometric with $P(X=x)=p(1-p)^{x-1}$ for $x \ge 1$ assuming you count the first success
To get the marginal distribution for $P(Y=y)$, this is then $\displaystyle \sum_{x=\max(1,y)}^{\infty} p(1-p)^{x-1} \times {x \choose y}p^y (1-p)^{x-y}$ which can be simplified.
For example $$P(Y=0) = \sum_{x=1}^{\infty} p (1-p)^{x-1} \times (1-p)^{x} \\ =p(1-p)\left(\sum_{x=0}^{\infty} (1-p)^{2x}-1\right)\\ = \dfrac{p(1-p)}{1-(1-p)^2} \\= \dfrac{1-p}{2-p} $$