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Hypergeometric Distribution Reference cheat sheet - grade college

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Statistics Grade college

Hypergeometric Distribution Reference Cheat Sheet

A printable reference covering hypergeometric probability, parameters, support, mean, variance, and sampling without replacement for college.

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The hypergeometric distribution models counts of successes in a fixed-size sample drawn without replacement from a finite population. Students need this reference because it separates hypergeometric settings from binomial settings, which is a common source of errors in statistics. It is especially useful for card problems, quality control, auditing, biology sampling, and survey selection.

The cheat sheet summarizes the model setup, probability formula, parameter restrictions, and interpretation of results.

The core idea is that a population of size NN contains KK successes and N−KN-K failures, and a sample of size nn is drawn without replacement. If XX is the number of successes in the sample, then X∼Hypergeometric⁡(N,K,n)X \sim \operatorname{Hypergeometric}(N,K,n). The probability mass function is P(X=x)=(Kx)(N−Kn−x)(Nn)P(X=x)=\frac{\binom{K}{x}\binom{N-K}{n-x}}{\binom{N}{n}}, with xx restricted to feasible values.

The mean is E(X)=nKNE(X)=n\frac{K}{N}, and the variance is Var⁡(X)=nKN(1−KN)N−nN−1\operatorname{Var}(X)=n\frac{K}{N}\left(1-\frac{K}{N}\right)\frac{N-n}{N-1}.

Key Facts

  • The hypergeometric distribution applies when sampling nn items without replacement from a finite population of size NN containing KK successes.
  • If XX counts successes in the sample, write X∼Hypergeometric⁡(N,K,n)X \sim \operatorname{Hypergeometric}(N,K,n).
  • The probability mass function is P(X=x)=(Kx)(N−Kn−x)(Nn)P(X=x)=\frac{\binom{K}{x}\binom{N-K}{n-x}}{\binom{N}{n}}.
  • The support is max⁡(0,n−(N−K))≤x≤min⁡(n,K)\max(0,n-(N-K)) \le x \le \min(n,K), where xx must be an integer.
  • The mean is E(X)=nKNE(X)=n\frac{K}{N}.
  • The variance is Var⁡(X)=nKN(1−KN)N−nN−1\operatorname{Var}(X)=n\frac{K}{N}\left(1-\frac{K}{N}\right)\frac{N-n}{N-1}.
  • The factor N−nN−1\frac{N-n}{N-1} is the finite population correction that makes the variance smaller than the binomial variance.
  • For large NN with small sampling fraction nN\frac{n}{N}, a binomial approximation with p=KNp=\frac{K}{N} is often reasonable.

Vocabulary

Population size
The total number of objects in the finite population, usually denoted by NN.
Success states
The number of population objects that have the characteristic being counted, usually denoted by KK.
Sample size
The number of objects drawn from the population without replacement, usually denoted by nn.
Support
The set of possible integer values of XX, from max⁡(0,n−(N−K))\max(0,n-(N-K)) to min⁡(n,K)\min(n,K).
Probability mass function
A formula that gives P(X=x)P(X=x) for each possible value of a discrete random variable.
Finite population correction
The multiplier N−nN−1\frac{N-n}{N-1} that adjusts variance because draws are made without replacement.

Common Mistakes to Avoid

  • Using the binomial distribution automatically is wrong because hypergeometric trials are not independent when sampling is without replacement.
  • Letting xx take impossible values is wrong because xx must satisfy max⁡(0,n−(N−K))≤x≤min⁡(n,K)\max(0,n-(N-K)) \le x \le \min(n,K).
  • Forgetting the failure term (N−Kn−x)\binom{N-K}{n-x} is wrong because a complete sample includes both selected successes and selected failures.
  • Using KK and nn interchangeably is wrong because KK is the number of successes in the population, while nn is the number of draws.
  • Computing variance as np(1−p)np(1-p) is wrong for an exact hypergeometric model because it omits the finite population correction N−nN−1\frac{N-n}{N-1}.

Practice Questions

  1. 1 A box contains 2020 parts, of which 66 are defective. If 55 parts are selected without replacement, find P(X=2)P(X=2) where XX is the number of defective parts.
  2. 2 A committee of 44 people is chosen from 1212 students, including 55 seniors. Find the probability that exactly 33 committee members are seniors.
  3. 3 For X∼Hypergeometric⁡(N=50,K=8,n=10)X \sim \operatorname{Hypergeometric}(N=50,K=8,n=10), compute E(X)E(X) and Var⁡(X)\operatorname{Var}(X).
  4. 4 Explain why drawing 1010 cards from a standard deck and counting hearts is modeled by a hypergeometric distribution rather than a binomial distribution.

Understanding Hypergeometric Distribution Reference

The important feature is dependence between draws. Once one item is selected, the population changes. If a batch has defective items and the first item drawn is defective, there is one fewer defective item left for the next draw.

The chance on the next draw is therefore different from the chance before. This is why multiplying the same success probability repeatedly gives the wrong model.

A tree diagram can show this changing probability clearly for small populations. For larger populations, counting groups of items is much more efficient than listing every possible draw order.

The probability rule comes from comparing equally likely samples. The denominator counts every possible group of the chosen sample size from the full population. The numerator counts only groups containing the required number of successes.

It chooses the successes first, then chooses the remaining sample members from the non-successes. Order does not matter here. Drawing red, blue, red produces the same sample group as blue, red, red.

Students often make errors by counting ordered sequences in one part of a calculation but unordered groups in another. Pick one method and use it consistently.

Feasible values deserve careful attention before any calculation. A sample cannot contain more successes than exist in the population. It cannot contain more successes than the sample size.

Sometimes it must contain a certain number of successes because there are too few non-successes available to fill the sample. For example, if an inspector selects eight products from a lot with only three non-defective products, at least five selected products must be defective. Checking these limits can quickly reveal an impossible answer, even when a calculator returns a number after incorrect input.

The average count is useful, but it is not a guarantee for one sample. If the expected number of defective products is two, a particular inspection can still find zero, one, or several. The spread is smaller than it would be for independent draws because removing items limits how extreme later results can become.

This matters in audits, warehouse checks, card hands, ecological capture studies, and classroom activities involving a bag of colored counters. A binomial model can be close when the sample is tiny compared with the population, since removal barely changes the proportions.

When the sample is a noticeable share of the population, use the hypergeometric model. Always identify the whole population, define exactly what counts as a success, and confirm that each item has an equal chance of selection.