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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.

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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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  • $\begingroup$ $b_2w_1b_2$ must be $b_2w_1w_2$. $\endgroup$ – farruhota Oct 19 '19 at 16:10
  • $\begingroup$ @farruhota: Typo corrected, thanks a lot. $\endgroup$ – Markus Scheuer Oct 19 '19 at 16:11
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I usually imagine the experiment as picking up $n$ balls from the bag of $N$ balls at once. In this way, one can't number balls as $ball1$, $ball2$ and so on. Every $n$ ball will have equal probability of getting picked then I guess.

Now if we want $r$ picks to be a success (occurrence of red ball in bag of $R$ red and $N-R$ blue balls, say) then we have to pick $r$ balls from $R$ and remaining $n-r$ balls from $N-R$.

Probability=$$\frac{\binom{R}{r}\binom{N-R}{n-r}}{\binom{N}{n}}$$

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all ${w + b}\choose{n}$ samples are equally likely...

Note that the dots will continue as, for example, the probability of $k$ white balls is $\frac{{w\choose k}{b\choose n-k}}{{w+b\choose n}}$.

Note that the samples have equal chance to be selected, however, to calculate the probability of $k$ white balls one needs to count the number of favorable samples and divide it by the total number of samples.

Example. Let's say you know there are $5$ balls in a box, but you do not know that $2$ are white and $3$ are black. You take out $4$ balls at random without replacement. If we label the balls as $1,2,3,4,5$, then there are ${5\choose 4}$ samples of balls to take out (order is not important): $$\{1,2,3,4\},\{1,2,3,5\},\{1,2,4,5\},\{1,3,4,5\},\{2,3,4,5\}$$ Do you agree that each sample is equally likely to be selected?

And now I ask a question: what is the probability that $1$ of the $4$ taken balls is white. Imagine that the balls numbered $1$ and $3$ were white. Then the question is to find the chance of the favorable samples occurring out of all possible samples. $2$ samples (the third and last) have $1$ or $3$, but not both (others). Hence, the probability of $1$ white in $4$ taken balls is $\frac25$. Indeed, the above formula confirms it: $$\frac{{w\choose k}{b\choose n-k}}{{w+b\choose n}}=\frac{{2\choose 1}{3\choose 4-1}}{{3+2\choose 4}}=\frac{2}{5}.$$

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