the probability that p takes value p and the number of favourable responses
is y. Now Y is a binomially distributed RV with probability of a fav. response for
an individual patient p, and number of patients n, so:
p(y,n|p)=n!/[(n-y)!y!)] p^y (1-p)^(n-y)
The unconditional (marginal?) distribution of Y is:
g(y,n)=int(0,1) P(y,n,p) dp
E(Y|n=2)= sum(y=0,infty) g(y,2)(B) Find E(Y) for n = 2.