Suppose I have a volume $V$ (ml) of a solution, containing a fixed number $n$ of particles of a certain substance (included in the volume count). Assume also we don't know anything about the particles spatial distribution in the volume.
If I take a random sample of volume $V_{sample}$ from my solution, and call $Y$ the number of substance particles in the sample, how can I compute the probability distribution of $Y$?
And what if the original number $n$ is not deterministic but itself distributed with some known pmf $P(n=i)$?
Since you're asking this on the math site and not on the physics site, I'm assuming that you're abstracting from physical effects like excluded volume and are assuming an ideal solution in which each solvent particle is independently uniformly distributed over the entire accessible volume.
Each solvent particle is in the sample volume $V_{\text{sample}}$ with independent probability $p=V_{\text{sample}}/V$. Then the number $Y$ of solvent particles in the sample volume $V_{\text{sample}}$ is binomially distributed with parameters $n$ and $p$:
$$ \mathsf P(Y=y)=\binom nyp^y(1-p)^{n-y}=\binom ny\left(\frac{V_{\text{sample}}}V\right)^y\left(1-\frac{V_{\text{sample}}}V\right)^{n-y}\;.$$
For the large numbers of particles contained in typical volumes, this binomial distribution is very well approximated by a normal distribution with the same mean and variance. The mean is $\mu=np=n\frac{V_{\text{sample}}}V$, and the variance is $\sigma^2=np(1-p)=n\frac{V_{\text{sample}}}V\left(1-\frac{V_{\text{sample}}}V\right)$.
If the number of particles is itself a random variable $N$, you need to apply the law of total probability:
$$ \mathsf P(Y=y)=\sum_{n=y}^\infty\mathsf P(N=n)\mathsf P(Y=y\mid N=n)\;, $$
where the conditional probability $\mathsf P(Y=y\mid N=n)$ is given by the expression for $\mathsf P(Y=y)$ in the first part of the answer.