Finding $P(X=0)$ given that we only have conditional mass function

Question

Let $Y$ be uniform on $(0,1)$ and $X|Y=y$ be binomial with paremeters $(n,y)$. Im looking to find $P(X=0)$

What i tried: well, we given that

$$ p(X|Y=y) = { n \choose x } y^x (1-y)^{n-y} $$

Now we know $E(X) = E (E(X|Y))=E(ny) = \frac{n}{2} $

Since $Y$ uniform so mean is $\frac{1}{2}$. Now we also know that

$$ \sum x P(X=x) =\frac{n}{2} $$

But here I get stuck since the requird probability we want gets multiplied by zero on left hand side. Now, am I approaching this problem correctly?

Answer 1

$\begin{split}\mathsf P(X=x)&=\int_0^1 \mathsf P(X=x\mid Y=y)f_Y(y)\mathsf d y\\ &= \binom nx \int_0^1 y^x(1-y)^{n-x}\mathsf d y\\\mathsf P(X=0)&= \binom n0\int_0^1(1-y)^n\mathsf d y\\ &= \ldots\end{split}$

Answer 2

Guide:

Try to evaluate the following integral\begin{align} P(X=0) &= \int_0^1 P(X=0|Y=y) \, dy \\ &=\int_0^1 (1-y)^n \, dy \end{align}