N: P(N=n)=(3/5)[(2/5)^n], find E(N)

Suppose the number of children N of a random family satifies P(N=n)=(3/5)[(2/5)^n] where n=0,1,2,... Compute E(N).

I went through the math and I come up with 1, which is the probability space. So, I know I'm not going about this the right way.

Could somone point out the basic understanding of expected value that I seem to be missing in this problem?

Re: N: P(N=n)=(3/5)[(2/5)^n], find E(N)

I'm going to assume you mean expected value. Here is the definition

What you have is the sum of all possible probabilities, which of course must be equal to 1!

However, each of your probabilities P[N=i] has a *weight* n_{i}

Re: N: P(N=n)=(3/5)[(2/5)^n], find E(N)

At least I was right about missing a basic understanding. The unfortunate part is that I need to learn how to go from the summation to the answer using paper and pencil, I won't always have access to a calculator that can do the sum for me...

Re: N: P(N=n)=(3/5)[(2/5)^n], find E(N)

I'm trying to remember back in my statistics class. If you were to take the natural log of both sides, would that help? when I see the product of n and (2/5)^n, that's what I want to do. EDIT:// but it is a sum still so it might not.