Computing absolute distribution from conditioned probability

Question

for a sum $X:= \sum_{i=1}^n 1_{U_n<U_0}$ of a series of random variables $U_1, ..., U_n$, all of them uniformely distributed on the unit interval as is $U_0$, I computed the following conditional probability:

$\mathbb{P}[X=k|U_0] = \binom{n}{k}U_0^k\cdot(1-U_0)^{n-k}$

Now I want to compute the absolute distribution of X, $\mathbb{P}[X=k]$. How do I do it? Please explain!

Answer 1

By definition of the conditional probability one has $$ P(X=k)={\rm E}[P(X=k\mid U_0)] $$ or equivalently $$ P(X=k)=\int_\mathbb{R}P(X=k\mid U_0=u) f_{U_0}(u)\,\mathrm du. $$