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?
 A: 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
A: Lets say we partition the container into several volumes: $V_i$ (i.e., $\sum V_i=V_0$)
At any time, we know that the number of molecules in each volume is some random number $N_i$, but $\sum N_i = N_0$
However, each molecule can be anywhere, therefore, this box will have uniform density $\delta$. This implies that $N_i\propto V_i$ or $N_i = kV_i$
Substituting this into the particle count constriant we get:
$$\sum kV_i=kV_0=N_0 \implies k=\frac{N_0}{V_0}$$
By linearity of expectation:
$$E\left[\sum \frac{N_0V_i}{V_0}\right]=\sum E\left[\frac{N_0V_i}{V_0}\right]=N_0 \implies E[N_i]=\frac{N_0V_i}{V_0} $$
Statistically speaking:
You can associate a bernoulli random variable with each particle: $B_i\sim Ber\left(\frac{V}{V_0}\right)$ Thus the number of particles in your box will have a binomial distribution:
$$N_i\sim Bin\left(N_0,\frac{V_i}{V_0}\right)$$
The expected value of a binomial distribution $Bin(n,p)$ is $np$. In your case:
$$E[N_i]=\frac{N_0V_i}{V_0}$$
A: Think of the problem this way:  each particle is a "trial" and the outcome of that trial is "success" if we find that particle inside the volume $V$.  The probability of any given trial (particle) being found inside $V$ is $p = V/V_0$.  Because the particles do not interact and are indistinguishable, then these trials are independent and identically distributed.  So the number of successes $X$ in $n = N_0$ trials is a binomially distributed random variable with probability of success $p$, and the expected number of such successes is simply $\operatorname{E}[X] = np = N_0 V/V_0$.
This analogy also gives us the exact probability distribution for the number of particles found in $V$:  it is $$\Pr[X = x] = \binom{N_0}{x} \left(\frac{V}{V_0}\right)^x \left(1 - \frac{V}{V_0}\right)^{N_0 - x}, \quad x = 0, 1, 2, \ldots, N_0.$$
We can prove that the expectation is $np$ by seeing that a binomial random variable is the sum of $n$ independent and identically distributed Bernoulli trials.  That is to say, if particle $i$ (where $i \in \{1, 2, \ldots, N_0\}$) is found in the volume $V$, then $X_i = 1$ with probability $p = V/V_0$, otherwise $X_i = 0$.  Then the expected value of $X_i$ is just $\operatorname{E}[X_i] = \Pr[X_i = 1] = p$, and the sum of $N_0$ such Bernoulli variables is $N_0 p = N_0 V/V_0$, as claimed.
