What is the main difference between the Hypergeometric and Binomial distributions?

The key difference is the method of sampling. The Binomial distribution assumes sampling *with replacement*, meaning the probability of success (p) is constant for every trial. The Hypergeometric distribution describes sampling *without replacement*, so each time an item is drawn, it is not returned, and the probability of success changes for the subsequent draws.

When should I use the Hypergeometric distribution?

Use the Hypergeometric distribution when you are taking a sample from a relatively small, finite population and you are not replacing the items after you draw them. A common rule of thumb is to use it when your sample size (n) is more than 5% of the total population size (N).

What is a real-world example of the hypergeometric distribution?

A classic example is a lottery. Imagine a lottery with 49 balls, and you need to match 6 winning balls. If you buy one ticket (a sample of 6 balls), the Hypergeometric distribution can calculate the probability of matching exactly 3 of the 6 winning balls. Here, N=49 (total balls), K=6 (winning balls), n=6 (your ticket), and you want to find the probability for k=3 (matching 3).

Hypergeometric Distribution Calculator (with Formula)

How to Use This Calculator

Step 1: Enter Your Parameters

Population Size (N): Total number of items in your population (e.g., 50 balls in a lottery)
Successes in Population (K): Number of "success" items in the population (e.g., 10 winning balls)
Sample Size (n): How many items you're drawing without replacement (e.g., 15 balls drawn)
Point 'a': Used to calculate P(X = a) and P(X < a) — the exact probability and cumulative probability up to this point
Point 'b': Used to calculate P(X > b) and P(a ≤ X ≤ b) — probability above this point and between a and b

Step 2: Click "Calculate"

The calculator will instantly show you:

Key probabilities: P(X = a), P(X < a), P(X > b), and P(a ≤ X ≤ b)
Statistical measures: mean, variance, and standard deviation
Visual charts: PMF (bar chart) and CDF (line chart)
Complete probability table for all possible values

Step 3: Export Your Results

Use the buttons below the results to:

Copy Table Data: Copy probability values to paste into Excel or other software
Save as PNG: Download charts and summary as an image
Save as PDF: Generate a complete report with all calculations and charts

💡 Quick Tip: The calculator automatically determines the valid range for k (number of successes in your sample). If you enter values for 'a' or 'b' outside this range, you'll see a warning, but the calculation will still proceed.

Understanding the Hypergeometric Distribution

The Hypergeometric Distribution is a discrete probability distribution that describes the probability of $k$ successes in a sample of size $n$ drawn without replacement from a finite population of size $N$ that contains exactly $K$ objects with a certain feature (successes).

Difference from Binomial Distribution

The key distinction is the "without replacement" condition. In a Binomial distribution, trials are independent, and the probability of success is constant. In a Hypergeometric distribution, each draw is dependent on the previous ones because the population size decreases, changing the probability of success for subsequent draws.

Use Binomial for sampling with replacement, or if the population is very large compared to the sample (e.g., sample size is < 5% of population size).
Use Hypergeometric for sampling without replacement from a small, finite population.

The Hypergeometric Formula

The probability of getting exactly $k$ successes is given by the Probability Mass Function (PMF):

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

Where:

$\binom{K}{k}$ is the number of ways to choose $k$ successes from the $K$ available successes in the population.
$\binom{N-K}{n-k}$ is the number of ways to choose the remaining $n-k$ items (failures) from the $N-K$ failures in the population.
$\binom{N}{n}$ is the total number of ways to choose a sample of size $n$ from the entire population $N$.

Mean / Expected Value: $E[X] = n \frac{K}{N}$

Variance: $Var[X] = n \frac{K}{N} (1-\frac{K}{N}) \frac{N-n}{N-1}$. The last term, $\frac{N-n}{N-1}$, is the finite population correction factor.

Examples with Step-by-Step Answers

Example 1: Quality Control

Problem: A batch of 20 items contains 5 defectives. If you randomly select 4 items without replacement, what is the probability that exactly 1 of them is defective?

1. Identify the parameters:

Population size, $N = 20$
Successes in population (defectives), $K = 5$
Sample size, $n = 4$
Successes in sample, $k = 1$

2. Apply the formula:

$$P(X=1) = \frac{\binom{5}{1} \binom{20-5}{4-1}}{\binom{20}{4}} = \frac{\binom{5}{1} \binom{15}{3}}{\binom{20}{4}}$$

3. Calculate the combinations: $\binom{5}{1}=5$, $\binom{15}{3}=455$, $\binom{20}{4}=4845$.

4. Combine the results: $P(X=1) = \frac{5 \times 455}{4845} = \frac{2275}{4845} \approx 0.4696$

Answer: There is a 46.96% chance of selecting exactly 1 defective item.

Example 2: Card Game

Problem: From a standard 52-card deck, you are dealt a 5-card hand. What is the probability that your hand contains exactly 2 aces?

1. Identify the parameters:

Population size (cards in deck), $N = 52$
Successes in population (aces), $K = 4$
Sample size (cards in hand), $n = 5$
Successes in sample, $k = 2$

2. Apply the formula:

$$P(X=2) = \frac{\binom{4}{2} \binom{52-4}{5-2}}{\binom{52}{5}} = \frac{\binom{4}{2} \binom{48}{3}}{\binom{52}{5}}$$

3. Calculate the combinations: $\binom{4}{2}=6$, $\binom{48}{3}=17296$, $\binom{52}{5}=2598960$.

4. Combine the results: $P(X=2) = \frac{6 \times 17296}{2598960} = \frac{103776}{2598960} \approx 0.0399$

Answer: The probability of getting exactly 2 aces in a 5-card hand is about 3.99%.

Hypergeometric Distribution Calculator

Results