How Do You Find the Unconditional Probability Distribution of Y?

[JaysFan31]

I'm in a university probability class studying multivariate distributions and have a problem I'm stuck on.

Here goes:
In a clinical study of a new drug formulated to reduce the effects of arthritis, researchers found that the proportion p of patients who respond favourably to the drug is a random variable that varies from batch to batch of the drug. Assume that p has a probability density function given by
f(p)={12*(p^(2))*(1-p), whenever p is between 0 and 1 inclusive
{0, whenever p is elsewhere.
Suppose that n patients are injected with portions of the drug taken from the same batch. Let Y denote the number showing a favourable response.
(A) Find the unconditional probability distribution of Y for general n.
(B) Find E(Y) for n = 2.

I'm confused because there's no Y1 and Y2. Every problem I've done has Y1 and Y2. How do you find the unconditional probability distribution for just Y in this case? I would love any help. Just a suggestion needed.

 

[HallsofIvy]

The proportion of patients who respond favorably to the drug is, by DEFINITION, the number of patients who responded favorably divided by the total number of patients injected with the drug.
If n patients are injected with the drug and the proportion who respond favorably is f(p) then the NUMBER who respond favorably is
nf(p), of course!

 

[JaysFan31]

Thanks for the response. I need to integrate the function from 0 to 1 I presume? What's E(Y) though?

 

[HallsofIvy]

Since Y= nf(p), E(Y) is the integral of that, from 0 to 1.