Choosing Between Hypergeometric and Binomial Models

Why the Sampling Design Determines the Model

A count of successes can look like the same random variable in two different studies, but the way the observations are generated matters. If a fixed number of distinct objects is selected from a finite population, the number of successes is generally modeled by a hypergeometric distribution. If there are a fixed number of independent trials with the same success probability on every trial, the count is modeled by a binomial distribution.

The key question is not simply whether the sample is “small” or whether the success probability seems nearly constant. First describe the design: Is each selected object returned before the next draw? Are outcomes independent? Is the success probability the same on every trial? Those details determine which model applies.

Definition: Suppose a population contains $N$ objects, of which $K$ are successes. A simple random sample of $n$ distinct objects drawn without replacement has a hypergeometric success count $X$, with parameters $N$, $K$, and $n$. Its probability mass function is $$ P(X=x)=\frac{\binom{K}{x}\binom{N-K}{n-x}}{\binom{N}{n}}, $$ for integer $x$ such that $\max(0,n-(N-K))\le x\le\min(n,K)$.

If $n$ independent trials each have the same success probability $p$, the success count $X$ has a binomial distribution with parameters $n$ and $p$, written $X\sim\operatorname{Binomial}(n,p)$, and $$ P(X=x)=\binom{n}{x}p^x(1-p)^{n-x},\qquad x=0,1,\ldots,n. $$

A Practical Model-Selection Check

In a hypergeometric design, the population has a fixed composition and the sample consists of distinct objects. A success changes what remains available: after drawing one success, there is one fewer success and one fewer total object. Thus the conditional success probability changes from draw to draw.

A binomial design has a fixed number of trials, independent outcomes, and a common success probability $p$. Drawing with replacement from a fixed population makes repeated draws independent, provided the population is restored and mixed before every draw. A binomial model also applies when distinct units have independent outcomes with the same probability, even if there is no physical replacement. Replacement is a common way to create independence, but it is not the definition of independence.

Specify the counted event.
Define what counts as a success and identify the random variable as the number of successes in the stated number of selections or trials.

Check for a fixed finite composition.
If there are $N$ distinct objects, exactly $K$ successes, and $n$ distinct objects are sampled without replacement, use the hypergeometric model.

Check independence and a common probability.
If there are $n$ independent trials and every trial has the same success probability $p$, use the binomial model.

Reject a familiar model if its assumptions fail.
If success probabilities differ across trials, or dependence is introduced by the design, neither standard model is justified without further argument.

A without-replacement sample from a finite population is not exactly binomial just because the population is large relative to the sample. The hypergeometric model remains the exact model under that design. Whether a binomial approximation is useful is a separate question and is not needed to select the exact model.

Why the Dependence Changes

The difference between the models can be seen by tracking the success indicators for two selections. In a simple random sample without replacement, let $I_1$ and $I_2$ equal 1 if the first and second objects are successes, and 0 otherwise. Write $p=K/N$, the success proportion in the population. Then

P(I_1=1,I_2=1)=\frac{K}{N}\frac{K-1}{N-1}.

The first factor is the probability that the first draw is a success. Given that it is, only $K-1$ successes remain among $N-1$ objects. Since each indicator has mean $p$, their covariance is

\begin{aligned} \operatorname{Cov}(I_1,I_2) &=E[I_1I_2]-E[I_1]E[I_2]\\ &=\frac{K}{N}\frac{K-1}{N-1}-\left(\frac{K}{N}\right)^2\\ &=-\frac{p(1-p)}{N-1}. \end{aligned}

For $0<K<N$, this covariance is negative: observing a success on the first draw makes a success on the next draw less likely. With replacement, both draws have success probability $p$ and are independent, so their covariance is zero. This dependence check gives a direct reason not to treat a without-replacement sample as an exact binomial experiment.

Worked Model Applications

Worked Example: Inspecting a Fixed Batch Without Replacement

A hypothetical crate contains 30 sensors, exactly 9 of which fail a calibration check. Four distinct sensors are selected at random without replacement. Let $X$ be the number that fail. Identify the model and calculate the probability of exactly two failures.

Method and assumptions. The population has $N=30$ objects and a fixed $K=9$ failures. The sample size is $n=4$, and no sensor is returned before the next selection. Therefore $X$ has a hypergeometric distribution.

For exactly two failures, choose 2 of the 9 failed sensors and 2 of the 21 passing sensors. The total number of equally likely samples of 4 sensors is $\binom{30}{4}$:

\begin{aligned} P(X=2) &=\frac{\binom{9}{2}\binom{21}{2}}{\binom{30}{4}}\\ &=\frac{36\cdot210}{27{,}405}\\ &=\frac{7{,}560}{27{,}405}\\ &\approx 0.2759. \end{aligned}

The probability is approximately $0.2759$, or 27.59%. The counts check: $\binom{9}{2}=36$, $\binom{21}{2}=210$, and $\binom{30}{4}=27{,}405$. The mean is $nK/N=4(9/30)=1.2$ failures, which is within the possible range from 0 to 4.

Common mistake avoided. Multiplying both the numerator and denominator counts by 10 would leave the ratio unchanged but would not make either combination count correct. Use the actual counts $7{,}560$ and $27{,}405$, not $75{,}600$ and $274{,}050$.

Worked Example: Independent Outcomes With a Common Probability

In a hypothetical administrative review, each of five independently selected files has probability $0.08$ of containing a specified coding error. Each file’s error status is independent of the others, and the probability is the same for all five. Let $X$ be the number of files with the error. Find the probability that exactly one file has it.

Method and assumptions. There are $n=5$ independent trials with common success probability $p=0.08$, where success means that a file contains the error. Thus $X\sim\operatorname{Binomial}(5,0.08)$.

Exactly one error can occur in any one of the five trial positions. The binomial formula gives

\begin{aligned} P(X=1) &=\binom{5}{1}(0.08)^1(0.92)^4\\ &=5(0.08)(0.71639296)\\ &=0.286557184\\ &\approx 0.2866. \end{aligned}

The probability is $0.2866$ to four decimal places. As a check on the setup, the binomial mean is $np=5(0.08)=0.4$ files. The mean is not itself the probability of exactly one error.

Common mistake avoided. Do not use a hypergeometric model just because a fixed number of files is reviewed. The stated outcomes are independent and do not come from a known fixed number of error-containing files in a finite batch. Those assumptions support the binomial model.

Worked Example: Drawing With Replacement and Using a Complement

A hypothetical container holds 10 tokens, 4 of which are green. A token is drawn, its color recorded, and it is returned and the container is mixed before the next draw. This is repeated three times. Let $X$ be the number of green draws. Find both $P(X=2)$ and the probability of at least one green draw.

Method and assumptions. Returning and remixing the token restores the same 4 green tokens among 10 before every draw. The draws are independent and each has $p=4/10=0.4$ probability of green, so $X\sim\operatorname{Binomial}(3,0.4)$.

For exactly two green draws, choose which two of the three positions are green. Each such pattern has probability $(0.4)^2(0.6)$:

\begin{aligned} P(X=2) &=\binom{3}{2}(0.4)^2(0.6)\\ &=3(0.16)(0.6)\\ &=0.288. \end{aligned}

For at least one green draw, use the complement of no green draws. Each draw has probability $0.6$ of not being green, so

\begin{aligned} P(X\ge 1) &=1-P(X=0)\\ &=1-(0.6)^3\\ &=1-0.216\\ &=0.784. \end{aligned}

The results are $P(X=2)=0.288$ and $P(X\ge1)=0.784$. For a check, $P(X=0)=0.216$, and $0.216+0.784=1$.

Common mistake avoided. Do not treat the second draw as having probability $3/9$ of green. That probability would apply after a green token was not replaced. Here the token is returned, so the green probability remains $4/10$ on each draw.

Common Model-Selection Traps

“Fixed sample size means binomial.” A fixed sample size occurs in both models. The decisive distinction is whether the trials are independent with a common probability or are distinct draws without replacement from a fixed composition.
“Without replacement means the binomial is close enough.” That may be a separate approximation question, but it does not change the exact model specified by the design. For a fixed finite population sampled without replacement, the exact count is hypergeometric.
“The success probability is the same on every draw.” Check the sampling rule. Without replacement, the conditional probability changes after each observed outcome, even though the unconditional success probability on any particular draw is $K/N$.
“Replacement is required for a binomial model.” Replacement from a fixed population is one route to independent trials. Independent outcomes on distinct units with the same probability also satisfy the binomial assumptions.
“A numerical match proves the model.” A formula can produce a plausible result even when its assumptions are wrong. State the design and justify the model before calculating.

Key takeaway: A hypergeometric model describes a fixed-size sample of distinct objects drawn without replacement from a population with a fixed number of successes. A binomial model describes a fixed number of independent trials with one common success probability. Select the model from the design, not from the appearance of the requested count.

Check Your Understanding

For each problem, identify the model where requested and calculate the stated probability. Answers are rounded to four decimal places when needed.

A hypothetical shipment has 24 distinct components, 6 of which are faulty. Three components are selected without replacement. What is the probability that none is faulty? (Answer: Hypergeometric; $\binom{18}{3}/\binom{24}{3}=816/2{,}024\approx0.4032$.)
A container has 8 tokens, 2 marked yellow. A token is drawn and replaced before each of four draws. What is the probability of exactly one yellow draw? (Answer: Binomial; $\binom{4}{1}(0.25)(0.75)^3=0.421875\approx0.4219$.)
Each of six independent checks has probability $0.10$ of detecting a specified issue. What is the probability that at least two checks detect it? (Answer: Binomial; $1-(0.9)^6-6(0.1)(0.9)^5=0.114265\approx0.1143$.)
A population contains 50 distinct devices, 10 of which have a feature. Five devices are sampled without replacement. Given that the first two sampled devices do not have the feature, what is the conditional probability that the third does? (Answer: Hypergeometric sampling; $10/48\approx0.2083$.)
Each of four independently selected hypothetical records has probability $0.12$ of containing an error, with the same probability for every record. What is the probability that none contains an error? (Answer: Binomial; $(1-0.12)^4=0.88^4=0.59969536\approx0.5997$.)

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