Confusion on how to calculate mean value I've done this type of thing in multiple classes over multiple years since high school, and still, when it's presented to me, I fumble around like a dope. 

Consider a gas of $N_0$ non-interacting molecules enclosed in
  a container of volume $V_0$. Focus attention on any sub-volume, $V$, of this
  container and denote by $N$ the number of molecules located within this
  sub-volume. Each molecule is equally likely to be located anywhere within
  the container; hence the probability that a given molecule is located within
  the sub-volume $V$ is simply equal to $\frac{V}{V_0}$.
(a) What is the mean number, $\bar{N}$, of molecules located within $V$? Express
  your answer in terms of $N_0$, $V_0$, and $V$.

So, the probability of a given molecule being in $V$ is just $\frac{V}{V_0}$. These molecules are independent of one another; thus, the probability of two given molecules being in $V$ should be $\left( \frac{V}{V_0} \right)^2$, the probability of three given molecules in $V$ should be $\left( \frac{V}{V_0} \right)^3$, etc. ... right?
The expected value of $N$ should be the sum of each outcome's probability multiplied by the outcome: $\bar{N} = \sum_{N = 1}^{N_0} \left( \frac{V}{V_0} \right)^NN$.
But this doesn't make sense; it should be the case that $\bar{N} \rightarrow N_0$ as $V \rightarrow V_0$. Clearly that does not happen in the above summation. I think, "Well, the probability is for a given particle. There are many ways that $V$ could have, say, 3 particles—there must be some issue with including duplicates." 
So, I divide each term in the summation by $N!$, since there are $N!$ ways of creating a particular result. $\bar{N} = \sum_{N = 1}^{N_0} \left( \frac{V}{V_0} \right)^N \frac{1}{(N-1)!}$. This is definitely not right.
Aside from all the different little things I've tweaked, by far, the biggest problem is that all of these things make complete and total sense to me. There's a very fundamental intuition that I have that's both inconsistent with other "intuitions" on the same thing, and very wrong.
Could someone offer some help?

The probability of ONLY two molecules being in the volume is
$p^2(1-p)^{N-2}$ times the combinatorial (n, 2). Where $p$ is V/V0.
Etc Etc
