# Probability of observing a number of special balls from a larger set

Suppose there is a collection of $$n$$ balls of which $$m$$ are special. What is the probability of drawing $$k$$ special balls, when $$p$$ balls are drawn?

To give it a try I considered the following particular case: Suppose there is $$n = 5$$ balls, of which $$m = 2$$ are special, and $$p = 4$$ balls are selected. The probability of observing $$1$$ special ball is then

$$(3/5) * (2/4) * (1/3) * 1 = 0.1$$

(first pick three non-special ones, then there is only special ones left). With the same logic, the probability of observing $$2$$ special balls would be

$$(3/5) * (2/4) * (2/3) * (1/2) = 0.1$$

which doesn't make sense to me (first pick two non-special, then two special). I don't see why the probability could be the same for observing $$1$$ or $$2$$ special ones.

How do I solve this problem?

Here, the number of special balls you draw, say $$X$$, is distributed according to hypergeometric distribution. And, according to its PMF definition, we have $$P(X=k)=\frac{{m\choose k}{n-m\choose p-k}}{{n \choose p}}$$
In the denominator, you count all possible $$p$$ drawings, and in the numerator you count all possible $$k$$ special drawings from $$m$$ balls together with the remaining $$p-k$$ possible non-special drawings from the remaining $$n-m$$ balls.