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?
Best Answer
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.