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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 NKN-K failures, and a sample of size nn is drawn without replacement. If XX is the number of successes in the sample, then XHypergeometric(N,K,n)X \sim \operatorname{Hypergeometric}(N,K,n). The probability mass function is P(X=x)=(Kx)(NKnx)(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(1KN)NnN1\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 XHypergeometric(N,K,n)X \sim \operatorname{Hypergeometric}(N,K,n).
  • The probability mass function is P(X=x)=(Kx)(NKnx)(Nn)P(X=x)=\frac{\binom{K}{x}\binom{N-K}{n-x}}{\binom{N}{n}}.
  • The support is max(0,n(NK))xmin(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(1KN)NnN1\operatorname{Var}(X)=n\frac{K}{N}\left(1-\frac{K}{N}\right)\frac{N-n}{N-1}.
  • The factor NnN1\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(NK))\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 NnN1\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(NK))xmin(n,K)\max(0,n-(N-K)) \le x \le \min(n,K).
  • Forgetting the failure term (NKnx)\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(1p)np(1-p) is wrong for an exact hypergeometric model because it omits the finite population correction NnN1\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 XHypergeometric(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.