How to calculate discrete probability distribution of number of particles in a sample? I have a statistical problem related to taking samples of liquid with microparticles in it. I have a solution of let’s say 1000 microparticles in 1 mL of water (large amount of this solution). The particles do not aggregate and are randomly distributed in the sample. I will take several small (1-2 µL) random samples from this solution. I’m aiming at having one particle in the sample but obviously only a part of the samples will have one particle.
The average amount of particles in the sample is very easy to calculate based on concentration and volume but what is not easy for me, is the distribution of number of particles in the samples. I’d like to achieve the maximum probability of having one particle/sample.
How is this distribution calculated? I guess we are talking about discrete probability distribution.
I’m very grateful for any help.
 A: I would start with a simpler problem - suppose you have ideal $U(0,1)$ random number generator, you want to sample $N$ numbers and wondering how many of them ($n$) will fit interval with length $l$ and what distribution would they follow.
If we denote probability to be in the interval with
$$
p = l
$$
and probability to be out of interval 
$$
q = 1-p = 1-l
$$
then you'll get binomial distribution with PMF
$$
PMF(N, n|l) = {N\choose n} l^n(1-l)^{N-n}
$$
Mean value would be $Nl$ which obviously consistent with $U(0,1)$ behavior. If you're looking at most probable number of samples in the interval, you shall look into mode of the distribution, which is either $floor((N+1)l)$ or $ceil((N+1)l)-1$. In your case
$$
l = \frac{v}{V}
$$
where $v$ is sample size, $V$ is whole volume size.
$$
N = \rho V
$$
where $\rho$ is concentration. Hope it helps...
A: *

*Suppose you take a sample with volume $x$ from your solution with total volume $X$. 

*If there's only one particle in the whole solution, the distribution of number of particles in your sample is $x/X$ (chance of having one particle) and $1-(x/X)$ (chance of having no particles).

*If there are multiple $(n)$ noninteracting particles, then each particle has the same probability $x/X$ of being in that first sample, and the same probability $1-(x/X)$ of not being in the first sample.  The number of particles in the sample follows a binomial distribution---that is, it behaves as if you flip $n$ unbalanced coins (with probability $x/X$ of heads), and the number of heads is the number of particles in your sample. That distribution on the number of particles $k$ is:
$$P(k) = {n \choose k}(x/X)^k (1-x/X)^{n-k}$$

*Based on properties of the binomial distribution, the average number of particles in the sample is $nx/X$.  The most likely number of particles is $\lfloor (n+1)x/X\rfloor$.  This means that if you want to most likely get one particle in your [first] sample, your sample size should be around $1/(n+1)$ of the total volume. Because of variance around this value, if you would prefer to err toward zero particles over multiple particles in your sample, you will want a smaller sample size than this.

*This math is for the first trial; in subsequent trials, the probability will technically depend on how many particles you've already fished out in previous samples.  (And also on the volume remaining in solution, if the samples are not negligibly small). This distribution is related to the hypergeometric distribution.

*However, if you take only a few, small samples from a comparatively large number of particles in a comparatively large total volume, then the number of particles $n$ and the total volume $X$ remains roughly constant even as you draw samples. So all samples have roughly the same probability distribution, and the $1/(n+1)$-or-less rule applies every time.
A: It seems reasonable to assume that the number of particles per sample follows a Poisson distribution.  If you have a sample size of $s$ microliters with a particle concentration of $1,000$ particles per milliliter, then the expected number of particles in the sample is simply $s$. Since we are assuming a Poisson distribution, the probability that there will be $k$ particles in the sample is 
$$\frac{s^k e^{-s}}{k!}$$
for $k = 0, 1, 2 , \dots$.
If we want to maximize the probability that $k=1$ then we want to maximize the function $s e^{-s}$ which, by standard methods of calculus, occurs at $s=1$.  So a sample size of $1$ microliter is best.
More generally, if you have a concentration of $c \times 1,000$ particles per milliliter, then the probability of getting exactly $k$ particles in a sample of size $s$ microliters is
$$\frac{(cs)^k e^{-cs}}{k!}$$
and the best sample size to get exactly one particle is $1/c$ microliters.
