The answer is E(CA)=EA*EB. This is why:
You have C=A*B. Therefore, E(CA)=E((A**2)*B))=E(A*B)=EA*EB.
The second to last equality holds because A**2=A, and the last one is
correct because A and B are independent.
Vadim
[EMAIL PROTECTED] (John Smith) wrote:
> If I have 3 variables defined as follo
; different balls. Then, E= SUM j*P{selected j different balls} over i=1
> to M1. Now, P{selected j different balls}= sub(N)Csub(j)*(j/N)^N,
> where sub(N)Csub(j) is the number of possible combinations of j items
> selected from N items, which is N!/(j!(N-j)!).
>
> Vadim Pliner
>
items, which is N!/(j!(N-j)!).
Vadim Pliner
[EMAIL PROTECTED] (Donald Burrill) wrote in message
news:<[EMAIL PROTECTED]>...
> On Sun, 26 Aug 2001 [EMAIL PROTECTED] wrote:
>
> > I have trouble to solve this probability problem. Hope get help here.
> >
> > There is
