Selecting a coin from from collection of unfair coins toss n times mean and probability of no tails We have a collection of unfair coins, for which the probability of getting head
after tossing them is uniformly distributed in the range (0, 1). We randomly select a coin
from this collection and toss it n times.
(a) What is the expected number of heads?
(b) Compute the probability of having no tails in these trials.
For part a) since its binomial I think it should be np but if we gonna use expectation of uniform on (0,1) which is 0.5 I don't know how to put it in
For b) i think it should be $1-{n \choose n}p^nq^{n-n}=1-p^n $

You have a free variable $p$ in your answer, but there's no such variable in the problem, so something must be wrong.
For (a), note that by symmetry the expected number of heads is equal to the expected number of tails, and these must add up to $n$, so they're both $n/2$.
For (b), you need to integrate over the distribution. If $p\in(0,1)$ is the probability of tossing heads, the probability of no tails (and thus all heads) is $p^n$. Integrating over the uniform distribution yields the probability
$$
\int_0^1p^n\mathrm dp=\frac1{n+1}\;.
$$
Note that this makes sense for $n=1$, where the probability must be $\frac12$ (again by symmetry).
