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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?

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1 Answer 1

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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.

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