The exact probability of observing $x$ unique elements each with property p after sampling with replacement from a set with elements

Question

Suppose I take $n$ random draws with replacement from a set of $N < \infty$ items. The probability that I will draw $k$ unique items is

$p(k) = \dfrac{(N)_k}{N^n} S(n, k)$

where $(N)_k = N (N-1) \cdots (N-k+1)$ is the "falling factorial" and $S(n, k)$ is the Stirling number of the second kind, i.e.,

$S(n, k) = \dfrac{1}{k!} \sum_{i=0}^k (-1)^i \binom{k}{i} (k-i)^n$.

This was solved previously here.

I am interested in an extension of this problem. Suppose further that $Q$ of the objects have a desirable property, and the remaining $N-Q$ objects do not. What then is the probability of drawing $k$ unique objects, where each object has property $Q$?

Thanks!

Answer 1

Let $Y$ denote the number of desirable objects extracted (counting duplicates), and $E_{N,n,k}$ be the event "to get $k$ diferent elements when drawing $n$ elements from $N$, with replacement".

Then the probability of our event $A$ is

$$ \begin{align} P(A)&=\sum_y P(A | Y=y) P(Y=y)\\ &=\sum_y P(E_{Q,y,k} ) P(Y=y)\\ &=\sum_{y=k}^n \dfrac{(Q)_k}{Q^y} S(y, k) \binom{n}{y} \frac{Q^y(N-Q)^{n-y}}{N^n}\\ \end{align}$$