Assume that X is uniformly distributed on (0, 1) and that the conditional distribution of Y given $X = x$ is a binomial distribution with parameters $(n, x)$. Then we say that Y has a binomial distribution with fixed size n and random probability parameter.
I have to find the marginal distribution of Y. So I have to use that: $$P(Y=y)=\sum_{x=y}^{\infty}P(Y=y|X=x)P(X=x)$$ But when we have that X is uniformly distributed I think the sum becomes a integral so we get: $$P(Y=y) =\int_0^1 \binom{n}{y} x^y(1-x)^{n-y}dx$$ But what to do next? I think maybe take the binomial coefficient out from the integral, but how then using this to find the marginal distribution of Y? Hope anyone can help me? Do I also have to use that Y has a binomial distribution with fixed size n and random probability parameter?
Your integral is correct. One can show that $$\int_0^1 x^y (1-x)^{n-y} \, dx = \frac{y! (n-y)!}{(n+1)!},$$ see the Beta function or order statistics of i.i.d. uniform random variables. Multiplying by $\binom{n}{y} = \frac{n!}{y! (n-y)!}$ and canceling gives a simple expression for $P(Y=y)$.