Hypergeometric Distribution: Concepts, Formulas, and Applications

Hypergeometric Distribution

Introduction to Hypergeometric Distribution

The hypergeometric distribution models the probability of obtaining a certain number of successes in a sample drawn without replacement from a finite population containing a fixed number of successes and failures. It is commonly used when sampling is done without replacement, making the probability of success change on each draw.

• Binomial Variable: Number of successes out of n trials, with constant probability p (sampling with replacement).

• Hypergeometric Variable: Number of successes in n draws from a group of N items, without replacement.

Comparing Binomial and Hypergeometric Distributions

Binomial Distribution	Hypergeometric Distribution
Sampling with replacement Each trial is independent Probability of success remains constant Formula: $P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$	Sampling without replacement Trials are dependent Probability of success changes after each draw Formula: $P(X = k) = \frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}}$

Key Terms and Definitions

• Population size (N): Total number of items.

• Number of successes in population (K): Number of items classified as 'success'.

• Sample size (n): Number of items drawn from the population.

• Number of observed successes (k): Number of successes in the sample.

Hypergeometric Probability Formula

The probability of getting exactly k successes in n draws from a population of N items with K successes is:

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

• $\binom{K}{k}$: Ways to choose k successes from K available.

• $\binom{N-K}{n-k}$: Ways to choose the remaining (n-k) failures from (N-K) available.

• $\binom{N}{n}$: Total ways to choose n items from N.

Example: Drawing Marbles

Problem: Take three marbles from a bag containing 2 red and 4 blue. Find the probability that exactly 1 of the marbles you pick is red (no replacement).

• N = 6 (total marbles)

• K = 2 (red marbles)

• n = 3 (marbles drawn)

• k = 1 (red marble drawn)

Apply the formula:

$P(X = 1) = \frac{\binom{2}{1} \binom{4}{2}}{\binom{6}{3}} = \frac{2 \times 6}{20} = \frac{12}{20} = 0.6$

Applications of the Hypergeometric Distribution

• Quality Control: Determining the probability of finding a certain number of defective items in a sample from a shipment.

• Lotteries and Raffles: Calculating the chance of winning when tickets are drawn without replacement.

• Card Games: Probability of drawing a specific combination of cards from a deck.

Worked Example: Quality Control

A quality control manager wants to ensure that their staff's testing procedure will identify defects in shipments reliably. Suppose a shipment contains 100 units, 5 of which are defective. The staff tests 20 units at random. What is the probability that at least 2 defective units are found in the sample?

• N = 100 (total units)

• K = 5 (defective units)

• n = 20 (units tested)

• k = 2 (at least 2 defective units found)

To find $P(X \geq 2)$, sum the probabilities for k = 2, 3, 4, 5 using the hypergeometric formula:

$P(X \geq 2) = \sum_{k=2}^{5} \frac{\binom{5}{k} \binom{95}{20-k}}{\binom{100}{20}}$

For k = 2:

$P(X = 2) = \frac{\binom{5}{2} \binom{95}{18}}{\binom{100}{20}}$

Continue similarly for k = 3, 4, 5 and sum the results.

Summary Table: Binomial vs. Hypergeometric

Feature	Binomial	Hypergeometric
Sampling	With replacement	Without replacement
Probability of Success	Constant	Changes after each draw
Independence	Independent trials	Dependent trials
Formula	$\binom{n}{k} p^k (1-p)^{n-k}$	$\frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}}$

Key Takeaways

• Use the hypergeometric distribution when sampling is done without replacement from a finite population.

• Use the binomial distribution when sampling is done with replacement or when the population is very large compared to the sample size.