Combinatorial reasoning behind Hypergeometric distribution

binomial-coefficientscombinatoricsprobability distributions

My textbook, Introduction to Probability by Blitzstein and Hwang, says the following when discussing the Hypergeometric distribution:

Story 3.4.1 (Hypergeometric distribution). Consider an urn with $w$ white balls and $b$ black balls. We draw $n$ balls out of the urn at random without replacement, such that all ${w + b}\choose{n}$ samples are equally likely. $\dots$

So the author says that we draw $n$ balls out of the urn at random without replacement, such that all ${w + b}\choose{n}$ samples are equally likely. I'm wondering what the combinatorial reasoning is behind the fact that all ${w + b}\choose{n}$ samples are equally likely? I'm familiar with what the binomial coefficient is, and I'm familiar with the concept of sampling without replacement (which is what the Hypergeometric distribution models), but I'm wondering if someone could provide a clear explanation of the combinatorial reasoning behind this.

I would greatly appreciate it if people could please take the time to clarify this.

Best Answer

Here we have a population of $w+b$ balls and we want to select $n$ balls at random without replacement. This means that each of the $\binom{w+b}{n}$ ways to select $n$ balls shall be equally probable.

Example: $w=3,b=2$ and $n=3$

We consider three white balls $w_1,w_2,w_3$ and two black balls $b_1,b_2$ in the urn. There are $$\binom{w+b}{3}=\binom{5}{3}=10$$ equally probable possibilities to draw three balls from the urn, namely

\begin{align*} &b_1b_2w_1\qquad \color{blue}{b_1w_1w_2}\qquad \color{blue}{b_2w_1w_2}\\ &b_1b_2w_2\qquad \color{blue}{b_1w_1w_3}\qquad \color{blue}{b_2w_1w_3}\qquad w_1w_2w_3\\ &b_1b_2w_3\qquad \color{blue}{b_1w_2w_3}\qquad \color{blue}{b_2w_2w_3}\\ \end{align*}

Since white balls are indistinguishible as are black balls, all possible configurations are listed above.

If we consider a hypergeometric distribution asking for the probability of drawing $2$ white balls in the example above (marked blue) we obtain \begin{align*} \frac{\binom{3}{2}\binom{2}{1}}{\binom{5}{3}} \end{align*}

with $\binom{5}{3}=10$ in the denominator indicating that each of the $10$ possibilities to draw three balls is equally probable.

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