---
title: "Hypergeometric distribution; binomial approximation"
subject: "Probability and Statistics"
chapter: 2
key: "semester_5/probability_and_statistics:hypergeometric-define-binomial-approx"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/hypergeometric-define-binomial-approx
site: BCT Engineering
---

# Hypergeometric distribution; binomial approximation

Probability and Statistics · Chapter 2: Discrete Probability Distribution

- Asked in: 2073 Shrawan, 5 marks
- Video: [Hypergeometric Distribution EXPLAINED!](https://www.youtube.com/watch?v=upVJ4YqTlC4) — zedstatistics
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/hypergeometric-define-binomial-approx
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-2

## Question

Define hypergeometric probability distribution with an example. Describe the conditions for the binomial approximation to hypergeometric distribution?

## Short answer

Hypergeometric distribution models successes in sampling without replacement from a finite population. Binomial approximation holds when population N is large, sample size n/N ≤ 0.05, and p=K/N not extreme; then probabilities are approximately binomial with p=K/N.

## Answer

#### Hypergeometric distribution
Suppose a finite population of N items has K "successes" (and N − K "failures"). If n items are selected at random **without replacement**, the number of successes X has the **hypergeometric distribution**:

P(X = x) = C(K, x) · C(N − K, n − x) / C(N, n), max(0, n − N + K) ≤ x ≤ min(n, K)

Mean = nK/N; variance = n (K/N)(1 − K/N)(N − n)/(N − 1).

**Example:** A box has 10 bulbs, 4 defective. Three are drawn without replacement. The probability that exactly 1 is defective is C(4,1) C(6,2)/C(10,3) = 4 × 15 / 120 = 0.5.

#### Binomial approximation to the hypergeometric
Draws without replacement become almost independent when the population is very large compared with the sample, so the hypergeometric can be replaced by B(n, p) with p = K/N:

C(K, x) C(N − K, n − x) / C(N, n) ≈ C(n, x) pˣ (1 − p)ⁿ⁻ˣ

Conditions:
1. The population size N is large.
2. The sample size is small compared with N, normally n/N ≤ 0.05 (some books say 0.1), so the finite population factor (N − n)/(N − 1) ≈ 1.
3. The proportion of successes p = K/N stays almost constant during sampling (K is also large, and p is not extremely close to 0 or 1).

Then mean nK/N ≈ np and variance ≈ npq.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What distinguishes the hypergeometric from the binomial?
   A. Sampling is with replacement
   B. Sampling is without replacement from a finite population
   C. The number of trials is random
   D. It counts rare events in time
   Answer: B. Draws without replacement change the success probability, so trials are not independent. (Key point: Defining feature)

2. P(X = x) in the hypergeometric is:
   A. C(K,x) C(N-K,n-x) / C(N,n)
   B. C(n,x) p^x q^(n-x)
   C. C(x-1,k-1) p^k q^(x-k)
   D. e^-lambda lambda^x / x!
   Answer: A. Choose x successes from K and n - x failures from N - K, over all samples. (Key point: Formula)

3. The binomial approximation is reasonable when:
   A. n/N is greater than 0.5
   B. N is small
   C. n/N is small (about 0.05 or less)
   D. n equals N
   Answer: C. A small sample fraction makes the finite population factor about 1. (Key point: Condition on sample size)

4. In the binomial approximation, p is taken as:
   A. n/N
   B. K/n
   C. (N-K)/n
   D. K/N
   Answer: D. The success proportion in the population. (Key point: Approximating p)

## Close variants set in other exams

- [Poisson definition; approximation to binomial](https://www.bctengineer.com/question.md?key=semester_5%2Fprobability_and_statistics%3Apoisson-definition-binomial-approx) — Probability and Statistics, chapter 2, asked 2×

## Related in Probability and Statistics

- [Binomial characteristics vs hypergeometric](https://www.bctengineer.com/question.md?key=semester_5%2Fprobability_and_statistics%3Abinomial-vs-hypergeometric) — Probability and Statistics, chapter 2, asked 4×
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- [Binomial definition and conditions](https://www.bctengineer.com/question.md?key=semester_5%2Fprobability_and_statistics%3Abinomial-definition-conditions) — Probability and Statistics, chapter 2
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