Hypergeometric Distribution Calculator
Calculate the probability of drawing a specific number of successes when sampling without replacement from a finite population, with a bar-chart diagram and step-by-step solution.
N = total population, K = successes in the population, n = sample size drawn without replacement, k = number of successes you want in that sample.
Hypergeometric Probability Distribution — P(X = 3)
The highlighted bars represent P(X = 3), which is approximately 29.79% when drawing a sample of n = 10 from a population of N = 50 containing K = 15 successes.
Step-by-Step Solution
Here's exactly how this answer was calculated, one step at a time.
- 1
Write down the given values
N = 50, K = 15, n = 10, k = 3
N is the population size, K is the number of successes in the population, n is the sample size drawn, and k is the exact number of successes wanted in the sample.
- 2
Count ways to choose k successes from K
C(K, k) = C(15, 3) = 455
This counts every possible way to pick k successes out of the K available in the population.
- 3
Count ways to choose the remaining items from the failures
C(N − K, n − k) = C(35, 7) = 6,724,520
The rest of the sample (n − k items) must come from the N − K non-success items in the population.
- 4
Count all possible samples of size n
C(N, n) = C(50, 10) = 10,272,278,170
This is the total number of ways to choose any sample of size n from the full population, ignoring successes or failures.
- 5
Divide to get the probability
P(X = 3) = [C(K,k) × C(N−K,n−k)] / C(N,n) = [455 × 6,724,520] / 10,272,278,170 = 0.2979
Dividing the favorable combinations by all possible combinations gives the exact probability.
✓ Final Answer: P(X = 3) ≈ 0.2979 (29.79%)
Free Online Hypergeometric Distribution Calculator
This hypergeometric distribution calculator finds the exact probability of drawing a specific number of successes when sampling without replacement from a finite population. Enter the population size, the number of successes in that population, the sample size, and the number of successes you want, and the calculator instantly returns the probability, mean, variance, and standard deviation — complete with a bar-chart diagram and a full step-by-step solution.
Whether you're a student solving probability problems for a statistics course, a quality-control engineer sampling a batch for defects, a researcher analyzing survey data, or a card-game enthusiast calculating the odds of a specific hand, this hypergeometric probability calculator handles the entire calculation for you, including P(X = k), P(X ≤ k), P(X ≥ k), and P(a ≤ X ≤ b).
Hypergeometric Distribution vs. Binomial Distribution
The hypergeometric distribution is easy to confuse with the binomial distribution, but there's one crucial difference: the hypergeometric distribution models sampling without replacement from a finite population, meaning each draw changes the odds for the next one — while the binomial distribution assumes each trial is independent with a constant probability of success (sampling with replacement, or from an effectively infinite population). If you're drawing cards from a deck without putting them back, or sampling items from a fixed batch without returning them, you need the hypergeometric distribution, not the binomial.
Hypergeometric Probability Formula
The probability of getting exactly k successes when drawing a sample of size n (without replacement) from a population of size N that contains K successes is given by the hypergeometric probability mass function:
P(X = k) = [C(K, k) · C(N − K, n − k)] / C(N, n)
Here, C(K, k) counts the ways to choose k successes from the K available in the population, C(N − K, n − k) counts the ways to fill the rest of the sample from the N − K non-success items, and C(N, n) counts every possible sample of size n from the full population, regardless of success or failure.
Mean and Variance of a Hypergeometric Distribution
Every hypergeometric distribution has a mean and variance that summarize its overall shape:
- Mean: μ = n·K / N (the expected number of successes in the sample)
- Variance: σ² = n·(K/N)·((N−K)/N)·((N−n)/(N−1))
- Standard Deviation: σ = √(variance)
- The (N−n)/(N−1) term is called the finite population correction factor, and it's what distinguishes the hypergeometric variance from the simpler binomial variance formula.
How to Use This Calculator
Select what you want to find — exactly k successes, at most k, at least k, or a probability between two values. Enter the population size (N), the number of successes in the population (K), the sample size (n), and the number of successes (k). The calculator instantly returns the exact probability, its percentage form, and the mean, variance, and standard deviation of the full distribution, along with a bar chart showing every possible outcome.
Worked Example
A batch of 50 items (N = 50) contains 15 defective items (K = 15). If you randomly sample 10 items without replacement (n = 10), what's the probability of finding exactly 3 defective items (k = 3)? Using the formula: P(X = 3) = [C(15,3) × C(35,7)] / C(50,10) = [455 × 6,724,520] / 10,272,278,170 ≈ 0.2979, or about 29.79%. The expected (mean) number of defective items in the sample is μ = 10 × 15/50 = 3.
Frequently Asked Questions
What is the hypergeometric distribution used for?
The hypergeometric distribution calculates probabilities when sampling without replacement from a finite population — for example, drawing cards from a deck, sampling defective items from a batch, or selecting a committee from a group without repetition.
What is the difference between hypergeometric and binomial distributions?
The binomial distribution assumes independent trials with a constant success probability (sampling with replacement). The hypergeometric distribution accounts for sampling without replacement from a finite population, where each draw changes the probability for the next.
What do N, K, n, and k mean in the hypergeometric formula?
N is the total population size, K is the number of successes in that population, n is the sample size you draw, and k is the number of successes you want to find within that sample.
What is the finite population correction factor?
It's the (N−n)/(N−1) term in the hypergeometric variance formula, which adjusts the variance downward compared to a binomial distribution, reflecting the fact that sampling without replacement reduces variability as the sample size approaches the population size.
Can the hypergeometric distribution be approximated by the binomial distribution?
Yes, when the population size N is much larger than the sample size n (a common rule of thumb is n/N < 0.05), sampling without replacement barely changes the odds between draws, so the binomial distribution becomes a good approximation.