The Hypergeometric Distribution

STA 211 · Lecture 10

Introduction

The hypergeometric distribution models sampling without replacement from a finite population containing two types of item. Unlike the binomial model, successive draws are dependent because the population composition changes after each draw.

Learning Objectives

You should be able to:

• recognise hypergeometric sampling;
• determine the correct support;
• calculate probabilities, means and variances;
• apply the finite-population correction; and
• compare hypergeometric and binomial models.

Probability Function

Suppose a population of size $N$ contains $K$ successes and $N - K$ failures. A sample of size $n$ is drawn without replacement. If $X$ is the number of successes in the sample, then

$$P(X = x) = \frac{\binom{K}{x}\binom{N - K}{n - x}}{\binom{N}{n}}.$$

The support is

$$max\{ 0,n - (N - K)\} \leq x \leq min\{ n,K\}.$$

Hypergeometric probability mass function

Hypergeometric probability mass function

Mean and Variance

Let

$$p = \frac{K}{N}.$$

Then

$$E(X) = np,$$

$$Var(X) = np(1 - p)\frac{N - n}{N - 1}.$$

The factor

$$\frac{N - n}{N - 1}$$

is the finite-population correction. It makes the hypergeometric variance smaller than the binomial variance $np(1 - p)$ because sampling without replacement contains less randomness.

Example 10.1

A batch of 20 components contains 5 defective items. A sample of 4 is selected without replacement. The probability of exactly 2 defects is

$$\frac{\binom{5}{2}\binom{15}{2}}{\binom{20}{4}} = \frac{1050}{4845} \approx 0.2167.$$

The mean is

$$E(X) = 4\left( \frac{5}{20} \right) = 1.$$

The variance is

$$4\left( \frac{1}{4} \right)\left( \frac{3}{4} \right)\left( \frac{16}{19} \right) \approx 0.6316.$$

Hypergeometric-to-Binomial Approximation

When the sampling fraction $n/N$ is small, the population composition changes very little and the draws are approximately independent. Then

$$X \approx Bin\left( n,\frac{K}{N} \right).$$

A common guide is that the approximation is reasonable when $n/N \leq 0.05$, though context and required precision matter.

Example 10.2: Model choice

From a population of 1,000 items with 100 defective items, a sample of 5 gives $n/N = 0.005$. A binomial approximation with $p = 0.1$ is usually accurate. From a population of 20 with 2 defective items and a sample of 10, the sampling fraction is 0.5; the exact hypergeometric model is essential.

Post-Test

1. State the difference between binomial and hypergeometric sampling.
2. A population of 30 contains 8 successes. A sample of 6 is selected without replacement. Find $P(X = 2)$.
3. For Question 2, find $E(X)$ and $Var(X)$.
4. Explain the finite-population correction.
5. When is a binomial approximation to the hypergeometric distribution reasonable?

Selected answers: 2. $\frac{\binom{8}{2}\binom{22}{4}}{\binom{30}{6}}$; 3. $1.6$ and $6(8/30)(22/30)(24/29) \approx 0.971$.

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