Lecture notes. Written from the material of MIT 6.041SC, licensed CC BY-NC-SA 4.0. These are notes, not a transcript: the material has been reorganised and rewritten. This adaptation carries the same licence, and the original is linked above.

26. Hypergeometric Probabilities¶

What this covers¶

This chapter answers a single counting question: an urn holds $n$ balls, $m$ of which are red and the rest black; we draw $k$ balls without replacement; what is the probability that exactly $i$ of the drawn balls are red? It assumes the reader already has the counting rule for combinations — $\binom{r}{j}$, the number of ways to choose $j$ objects out of $r$ — and the equally-likely-outcomes model of probability: when a sample space is a finite set of equally likely outcomes, the probability of an event is the fraction of the sample space it occupies.

The urn model¶

Picture the $n$ balls in a box, split by a line into two groups: $m$ red balls on one side and $n-m$ black balls on the other — "red" and "black" just name "the kind being counted" and "everything else." From this box a subset of $k$ balls is drawn: order doesn't matter, and a ball is not returned once drawn. Write $p_r$ for the probability that exactly $i$ of the $k$ drawn balls are red, so that $k-i$ are black.

The sample space¶

Since a draw is an unordered subset of $k$ balls chosen from $n$, the natural sample space is the set of all such subsets: $$\Omega = \{\text{all } k\text{-subsets of the } n \text{ balls}\}, \qquad |\Omega| = \binom{n}{k}.$$ Every $k$-subset is equally likely, so the probability of any event is the count of subsets belonging to it, divided by $\binom{n}{k}$.

Counting the subsets with exactly $i$ red balls¶

Let $c$ be the number of $k$-subsets containing exactly $i$ red balls. Building one such subset is a two-stage choice, and the stages don't interact: first pick which $i$ of the $m$ red balls are in the subset, then separately pick which $k-i$ of the $n-m$ black balls are in it. If $a$ counts the first stage and $b$ the second, the counting principle gives $c = a \cdot b$, and each stage is itself a combination count: $$a = \binom{m}{i}, \qquad b = \binom{n-m}{k-i}, \qquad c = \binom{m}{i}\binom{n-m}{k-i}.$$

The box of n balls split into m red and n − m black; the dashed ellipse is one draw of k balls, taking i from the red side and k − i from the black side.

The hypergeometric probability¶

Dividing the count of favorable subsets by the size of the sample space gives $$p_r = P(\text{exactly } i \text{ red}) = \frac{\binom{m}{i}\binom{n-m}{k-i}}{\binom{n}{k}}.$$ This is the hypergeometric probability — as a function of $i$, the hypergeometric distribution: the law governing the number of "special" items (here, red balls) in a fixed-size sample drawn without replacement from a finite population containing a fixed number of them.

Worked example: aces in a seven-card hand¶

Take a standard deck, $n = 52$ cards, of which $m = 4$ are aces, so $n - m = 48$ are not. Deal $k = 7$ cards and ask for the probability that exactly $i = 3$ of them are aces. Casting the deck into the urn picture — aces playing the role of red balls, the rest of the deck playing the role of black balls, dealing a hand playing the role of drawing $k$ balls — lets the formula carry over unchanged; only the substitution changes: $$p_r = \frac{\binom{4}{3}\binom{48}{4}}{\binom{52}{7}} = \frac{4 \times 194580}{133784560} \approx 0.0058.$$

Sources¶

Everything above is from the single supplied document for this chapter: a model-reconstructed transcript of an MIT 6.041SC recitation, "Hypergeometric Probabilities" (docs/probability/mit-ocw/6041sc/other/edit2-take2-no13-ch1-hypergeometicprobabilities-slides.md in the library repository), converted from a slide PDF that had no usable text layer. The source file itself flags that the prose is a paraphrase and every equation is unverified, reconstructed by a model reading the scanned pages — treat the formulas here as a pointer into the original slides, not as independently checked.
No separate slide deck, lecture transcript, written notes, or problem set were supplied alongside it for this chapter; it is a single stray recording, as reflected in the short length here.
The final decimal in the worked example ($\approx 0.0058$) completes the substitution the source sets up but stops short of evaluating; it is arithmetic on the source's own numbers, not an added example.

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