Hypergeometric Distribution

Deal $5$ cards and count the hearts. Pick a committee of $4$ from a club and count the Grade 12s. In both cases, once something is chosen it can’t be chosen again, so the trials are dependent and the binomial formula doesn’t apply. The hypergeometric distribution handles this kind of counting: sampling without replacement.

Key ideas

When is it hypergeometric?

A random variable $X$ has a hypergeometric distribution when:

• you choose $n$ items without replacement from a population of $N$ items;
• $a$ of the $N$ items are “successes” and the other $N - a$ are “failures”;
• $X$ is the number of successes you choose.

Because nothing is put back, the probability of a success changes from one pick to the next. That’s what makes the trials dependent.

The formula

$$P(X = k) = \frac{\dbinom{a}{k} \dbinom{N - a}{n - k}}{\dbinom{N}{n}}$$

This is just probability with counting:

• $\binom{N}{n}$ counts all the ways to choose $n$ items from $N$ (the sample space);
• $\binom{a}{k}$ counts the ways to choose $k$ of the successes;
• $\binom{N - a}{n - k}$ counts the ways to choose the remaining $n - k$ items from the failures.

The possible values of $k$ go from $0$ up to $n$, but you can’t choose more successes than there are ($k \le a$) or more failures than there are ($n - k \le N - a$). For impossible values, the formula gives $0$.

Technology: a spreadsheet has =HYPGEOM.DIST(k, n, a, N, FALSE) for $P(X = k)$. On a calculator without a hypergeometric function, compute the three combinations with nCr.

Expected value

$$E(X) = \frac{na}{N}$$

The fraction of successes in the population is $\tfrac{a}{N}$, so in a sample of $n$ you expect $n \times \tfrac{a}{N}$ of them. It’s the same idea as $E(X) = np$ for the binomial, with $p = \tfrac{a}{N}$.

Comparing with the binomial

	Binomial	Hypergeometric
Sampling	with replacement (or independent trials)	without replacement
Trials	independent	dependent
Probability of success	stays at $p$	changes after each pick
Expected value	$np$	$\dfrac{na}{N}$

Suppose $40\%$ of a population are successes and you choose $5$ items. The figure compares the two distributions for a small population and a large one.

Choosing $5$ items when $40\%$ of the population are successes. With $N = 20$ the two distributions differ; with $N = 400$ they’re nearly identical.

Two things to notice:

• Both distributions have the same expected value: $5(0.4) = 2$ and $\tfrac{5(8)}{20} = 2$.
• Without replacement, extreme results (like $0$ or $5$ successes) are less likely, so the hypergeometric histogram is a bit narrower and taller in the middle.

When the population is large compared with the sample, removing a few items barely changes the probability of success, so the hypergeometric and binomial distributions are almost the same. That’s why surveys of a few hundred people out of millions can be treated as binomial.

Worked examples

Example 1: A committee

A club has $12$ members: $7$ in Grade 12 and $5$ in Grade 11. A committee of $4$ is chosen at random. Let $X$ be the number of Grade 12 students on the committee. Make the probability distribution, draw a conclusion from its histogram, and find $E(X)$.

Solution. Here $N = 12$, $a = 7$, $n = 4$, and $\binom{12}{4} = 495$.

$$P(X = k) = \frac{\binom{7}{k} \binom{5}{4 - k}}{495}$$

$k$	$0$	$1$	$2$	$3$	$4$
Ways $\binom{7}{k}\binom{5}{4 - k}$	$1 \times 5 = 5$	$7 \times 10 = 70$	$21 \times 10 = 210$	$35 \times 5 = 175$	$35 \times 1 = 35$
$P(X = k)$	$0.0101$	$0.1414$	$0.4242$	$0.3535$	$0.0707$

Check: $5 + 70 + 210 + 175 + 35 = 495$. ✓

The histogram would have its tallest bar at $k = 2$, with $k = 3$ close behind. An all-Grade-11 committee ($k = 0$) is very unlikely: about $1\%$.

$$E(X) = \frac{na}{N} = \frac{4(7)}{12} = \frac{7}{3} \approx 2.33$$

Example 2: Hearts in a hand

Five cards are dealt from a standard deck. Find the probability of exactly $2$ hearts.

Solution. $N = 52$, $a = 13$ hearts, $n = 5$, $k = 2$.

$$P(X = 2) = \frac{\binom{13}{2}\binom{39}{3}}{\binom{52}{5}} = \frac{78 \times 9139}{2\,598\,960} = \frac{712\,842}{2\,598\,960} \approx 0.2743$$

Example 3: Quality control

A box of $20$ phone chargers contains $3$ defective ones. An inspector tests $4$ chargers chosen at random. Find the probability that at least one is defective.

Solution. Use the complement, as with the binomial. $N = 20$, $a = 3$, $n = 4$.

$$P(X = 0) = \frac{\binom{3}{0}\binom{17}{4}}{\binom{20}{4}} = \frac{2380}{4845} = \frac{28}{57}$$ $$P(X \ge 1) = 1 - \frac{28}{57} = \frac{29}{57} \approx 0.5088$$

Example 4: With or without replacement

In the figure above, $5$ items are chosen from a population that is $40\%$ successes. Find $P(X = 2)$ three ways: binomial, hypergeometric with $N = 20$ (and $a = 8$), and hypergeometric with $N = 400$ (and $a = 160$).

Solution.

Binomial (with replacement), $n = 5$, $p = 0.4$:

$$P(X = 2) = \binom{5}{2}(0.4)^2(0.6)^3 = 10 \times 0.16 \times 0.216 = 0.3456$$

Hypergeometric, $N = 20$:

$$P(X = 2) = \frac{\binom{8}{2}\binom{12}{3}}{\binom{20}{5}} = \frac{28 \times 220}{15\,504} \approx 0.3973$$

Hypergeometric, $N = 400$:

$$P(X = 2) = \frac{\binom{160}{2}\binom{240}{3}}{\binom{400}{5}} \approx 0.3478$$

With only $20$ items, sampling without replacement makes a big difference. With $400$ items, the answer is within $0.003$ of the binomial.

Common mistakes

Using the binomial formula for sampling without replacement. If items aren’t put back and the population is small, the trials are dependent. Use the hypergeometric formula.

Mixing up n and N, or a and k. $N$ is the population, $n$ the sample; $a$ is the number of successes in the population, $k$ the number in the sample. Write all four down before you start.

Choosing the failures from the wrong group. The second combination is $\binom{N - a}{n - k}$: the remaining $n - k$ items come from the $N - a$ failures. Check that the top numbers add to $N$ and the bottom numbers add to $n$.

Forgetting that some values are impossible. With only $3$ defective chargers, you can’t find $4$. Then $\binom{3}{4} = 0$, so $P(X = 4) = 0$.

Adding many terms for “at least one”. Just like the binomial, use $1 - P(X = 0)$.

Practice

1. (Warm-up) Binomial or hypergeometric?

• (a) $X$ is the number of girls when $3$ students are chosen at random from a class of $25$.
• (b) $X$ is the number of $4$s in $10$ rolls of a die.
• (c) $X$ is the number of red cards when $6$ cards are drawn from a deck, replacing each card before the next draw.

Solution

(a) Hypergeometric: students are chosen without replacement from a small class.

(b) Binomial: independent trials with $p = \tfrac{1}{6}$.

(c) Binomial: with replacement, $p = \tfrac{1}{2}$ every time.

2. (Warm-up) A bag has $6$ red and $4$ green marbles. Three are drawn without replacement. Find the probability that exactly $2$ are red.

Solution

$$P(X = 2) = \frac{\binom{6}{2}\binom{4}{1}}{\binom{10}{3}} = \frac{15 \times 4}{120} = \frac{60}{120} = \frac{1}{2}$$

3. (Warm-up) A sample of $10$ is chosen without replacement from a population of $50$ that includes $15$ successes. Find the expected number of successes.

Solution

$$E(X) = \frac{na}{N} = \frac{10(15)}{50} = 3$$

4. (Core) A science club has $10$ members, $4$ of them in Grade 9. Three members are chosen at random to go to a conference. Let $X$ be the number of Grade 9s chosen. Make the probability distribution of $X$ and find $E(X)$.

Solution

$N = 10$, $a = 4$, $n = 3$, and $\binom{10}{3} = 120$.

$k$	$0$	$1$	$2$	$3$
Ways $\binom{4}{k}\binom{6}{3 - k}$	$1 \times 20 = 20$	$4 \times 15 = 60$	$6 \times 6 = 36$	$4 \times 1 = 4$
$P(X = k)$	$\tfrac{1}{6} \approx 0.1667$	$\tfrac{1}{2} = 0.5000$	$\tfrac{3}{10} = 0.3000$	$\tfrac{1}{30} \approx 0.0333$

Check: $20 + 60 + 36 + 4 = 120$. ✓

$$E(X) = \frac{3(4)}{10} = 1.2$$

5. (Core) In a lottery, $6$ numbers are drawn from $1$ to $49$. You pick $6$ numbers. Find the probability that exactly $3$ of your numbers are drawn.

Solution

Think of the $6$ drawn numbers as the “successes” among $N = 49$. Your $6$ picks are the sample.

$$P(X = 3) = \frac{\binom{6}{3}\binom{43}{3}}{\binom{49}{6}} = \frac{20 \times 12\,341}{13\,983\,816} = \frac{246\,820}{13\,983\,816} \approx 0.0177$$

6. (Core) A pond has $30$ fish, and $8$ of them have been tagged. A biologist catches $5$ fish (without putting any back until the end). Find the probability that at least $2$ are tagged.

Solution

$N = 30$, $a = 8$, $n = 5$, and $\binom{30}{5} = 142\,506$. Use the complement.

$$P(X = 0) = \frac{\binom{8}{0}\binom{22}{5}}{142\,506} = \frac{26\,334}{142\,506} \approx 0.1848$$$$P(X = 1) = \frac{\binom{8}{1}\binom{22}{4}}{142\,506} = \frac{8 \times 7315}{142\,506} = \frac{58\,520}{142\,506} \approx 0.4106$$$$P(X \ge 2) = 1 - P(X = 0) - P(X = 1) \approx 0.4046$$

7. (Core) Five cards are drawn from a standard deck. Find the probability of exactly $2$ hearts if the cards are drawn with replacement. Compare with Example 2, and explain the difference.

Solution

With replacement, it’s binomial with $n = 5$, $p = \tfrac{13}{52} = 0.25$:

$$P(X = 2) = \binom{5}{2}(0.25)^2(0.75)^3 = 10 \times 0.0625 \times 0.421\,875 \approx 0.2637$$

Without replacement (Example 2), it’s about $0.2743$. The answers are close because $5$ cards is a small part of the $52$-card deck. Without replacement, the results bunch a little more tightly around the expected value ($1.25$ hearts): values close to it, like $1$ and $2$, become slightly more likely, and extreme values like $0$ or $5$ become slightly less likely.

8. (Challenge) Thirteen cards are dealt from a standard deck. Find the probability that the hand contains at least one ace.

Solution

$N = 52$, $a = 4$ aces, $n = 13$. Use the complement:

$$P(X = 0) = \frac{\binom{4}{0}\binom{48}{13}}{\binom{52}{13}} = \frac{6327}{20\,825} \approx 0.3038$$$$P(X \ge 1) = 1 - \frac{6327}{20\,825} = \frac{14\,498}{20\,825} \approx 0.6962$$

9. (Challenge) To estimate the number of fish in a lake, biologists catch, tag, and release $40$ fish. Later, they catch $50$ fish and find that $8$ are tagged. Use the expected value of a hypergeometric distribution to estimate the number of fish in the lake.

Solution

In the second catch, $n = 50$, $a = 40$ tagged fish, and $N$ is unknown. Assume the number of tagged fish caught equals the expected value:

$$\frac{na}{N} = 8 \quad\Rightarrow\quad \frac{50(40)}{N} = 8 \quad\Rightarrow\quad N = \frac{2000}{8} = 250$$

The lake has about $250$ fish. (This is called the capture–recapture method.)