Hypergeometric Versus Binomial Approximation

Why Compare These Two Count Models?

A count of successes in a sample can follow different probability models depending on how the sample is drawn. If the population is fixed and each selected item is not returned before the next selection, the draws are dependent and the exact count model is hypergeometric. A binomial model treats trials as independent, each with the same probability of success. It can be a useful approximation when the sample is small relative to the population, but it is not automatically accurate for every event.

The practical question is not simply whether the population is “large.” Check the sampling fraction, compare the event probabilities when feasible, and confirm that the binomial model does not assign probability to counts that the fixed population makes impossible. These are important model-selection checks in problems aligned with the SOA Exam P (Probability) syllabus.

Definition: Suppose a population contains $N$ items, of which $K$ are successes, and a simple random sample of size $n$ is drawn without replacement. If $X$ is the number of successes in the sample, then $X$ has a hypergeometric distribution: $$ P(X=k)=\frac{\binom{K}{k}\binom{N-K}{n-k}}{\binom{N}{n}}, \qquad \max(0,n-(N-K))\leq k\leq\min(n,K). $$ The population success proportion is $p=K/N$. The corresponding binomial approximation is $X_B\sim\operatorname{Binomial}(n,p)$, with $$ P(X_B=k)=\binom{n}{k}p^k(1-p)^{n-k},\qquad k=0,1,\ldots,n. $$

The hypergeometric formula counts ways to choose $k$ of the $K$ successes and $n-k$ of the $N-K$ failures, out of all samples of size $n$. The binomial model instead describes $n$ independent trials with success probability $p$. Its parameter $p$ matches the population fraction, but matching that parameter does not make the models identical.

Feasible Counts and the Sampling Fraction

For a fixed population, the count cannot exceed the number of successes available, $K$, or the sample size, $n$. It also cannot be so small that the sample would need more failures than the population contains. These constraints give the hypergeometric support: $ \max(0,n-(N-K))\leq X\leq\min(n,K) $. By contrast, a binomial count with $0<p<1$ has positive probability at every integer from 0 through $n$. When these support ranges differ, the binomial model assigns positive probability to some counts that are impossible in the fixed population. This matters especially for tail events involving those counts. The supports can also coincide; a support mismatch should be checked, not presumed.

The sampling fraction $n/N$ is a useful first diagnostic. A small fraction generally means that removing one item has little effect on the success fraction among the remaining items. Rules of thumb sometimes treat sampling at no more than about 5% of a population as a setting where the binomial approximation may be reasonable. This is only a heuristic, not a guarantee for every probability or tail event. When the fraction is larger, the dependence from sampling without replacement is more pronounced, so the exact model deserves particular attention.

Worked Example: Comparing an Exact Count With Its Approximation

A hypothetical inspection samples $n=5$ components without replacement from a fixed lot of $N=40$. Exactly $K=8$ components have a specified defect. Find the probability of exactly two defective components under the exact model and under the binomial approximation, then compare them.

Method and model. The exact count $X$ is hypergeometric, with $p=K/N=8/40=0.2$. The approximation is $X_B\sim\operatorname{Binomial}(5,0.2)$. The sample fraction is $5/40=0.125$, so the binomial result should be treated as an approximation to check, not as an automatic substitute.

Exact probability. There are $\binom{8}{2}=28$ ways to select two defective components and $\binom{32}{3}=4960$ ways to select three nondefective components. The total number of samples is $\binom{40}{5}=658008$. Therefore,

P(X=2)=\frac{\binom{8}{2}\binom{32}{3}}{\binom{40}{5}} =\frac{28(4960)}{658008} =\frac{138880}{658008} \approx 0.2111.

Binomial approximation. Using $n=5$, $p=0.2$, and $1-p=0.8$,

P(X_B=2)=\binom{5}{2}(0.2)^2(0.8)^3 =10(0.04)(0.512) =0.2048.

The absolute difference is approximately $0.2110613-0.2048=0.0062613$, or $0.0063$ to four decimal places. The exact probability is larger in this example. The calculation avoids the common mistake of assuming that use of the same success fraction makes the two event probabilities equal.

Why the Hypergeometric Variance Is Smaller

Both models have mean $np$, but sampling without replacement reduces the variance when the population has both successes and failures and more than one item is sampled. The reason is negative dependence: after a success is drawn, one fewer success remains available for later draws.

Result: For a hypergeometric count from a population with $N\geq1$, $K$ successes, and sample size $0\leq n\leq N$, let $p=K/N$. Then $$ E[X]=np,\qquad \operatorname{Var}(X)=np(1-p)\frac{N-n}{N-1} $$ when $N>1$. For $n=0$, the count is identically zero. The binomial approximation has mean $np$ and variance $np(1-p)$. For $0<p<1$ and $1\leq n\leq N$, the ratio of the hypergeometric variance to the binomial variance is $(N-n)/(N-1)$. The variances are equal when $n=0$, $n=1$, or $p=0$ or $p=1$; the hypergeometric variance is strictly smaller when $0<p<1$ and $n>1$.

Here is a justification of the variance formula. Write $X=I_1+\cdots+I_n$, where $I_j$ equals 1 if the $j$th draw is a success and 0 otherwise. Each draw has success probability $p$, so $E[I_j]=p$ and $\operatorname{Var}(I_j)=p(1-p)$. For two distinct draws, the probability that both are successes is $K(K-1)/(N(N-1))$. Thus their covariance is

\operatorname{Cov}(I_j,I_l) =\frac{K(K-1)}{N(N-1)}-p^2 =-\frac{p(1-p)}{N-1},\qquad j\ne l.

Adding the individual variances and all pairwise covariance terms gives the result:

\begin{aligned} \operatorname{Var}(X) &=n p(1-p)+n(n-1)\left(-\frac{p(1-p)}{N-1}\right)\\ &=np(1-p)\left(1-\frac{n-1}{N-1}\right)\\ &=np(1-p)\frac{N-n}{N-1}. \end{aligned}

The factor $(N-n)/(N-1)$ is called the finite-population correction to the variance. It quantifies the reduction in variance relative to the binomial model under the stated nondegenerate conditions. A variance comparison alone does not establish that every event probability is close: a particular tail or boundary event can be sensitive to the model even when the variances are similar.

Worked Example: A Large Sampling Fraction and an Impossible Tail

A hypothetical sample of $n=8$ items is drawn without replacement from a population of $N=20$, containing $K=5$ marked items. Find the probability of at least three marked items exactly and approximately. Explain why the support matters.

Method and model. Here $p=5/20=0.25$, and the sample fraction is $8/20=0.4$. The exact count is hypergeometric. Its feasible range is 0 through 5, because there are only five marked items in the population. The binomial approximation is $X_B\sim\operatorname{Binomial}(8,0.25)$, whose support extends through 8.

Exact probability. For $X\geq3$, the feasible counts are 3, 4, and 5. The denominator is $\binom{20}{8}=125970$. The three numerator terms are $\binom{5}{3}\binom{15}{5}=10(3003)=30030$, $\binom{5}{4}\binom{15}{4}=5(1365)=6825$, and $\binom{5}{5}\binom{15}{3}=1(455)=455$. Therefore,

P(X\geq3) =\frac{30030+6825+455}{125970} =\frac{37310}{125970} \approx 0.2962.

Binomial approximation. It is convenient to calculate the complement:

\begin{aligned} P(X_B\geq3) &=1-\left[P(X_B=0)+P(X_B=1)+P(X_B=2)\right]\\ &=1-\left[(0.75)^8+8(0.25)(0.75)^7 +\binom{8}{2}(0.25)^2(0.75)^6\right]\\ &=1-(0.1001129+0.2669678+0.3114624)\\ &\approx 0.3215. \end{aligned}

The approximation is higher by about $0.3214569-0.2961816=0.0252753$, or $0.0253$ to four decimal places. In this case the binomial support also includes counts 6, 7, and 8, which are impossible in the fixed population. Those extra outcomes are part of the approximate tail probability. The large sampling fraction is another reason not to rely on the approximation without checking it. The common trap is to include impossible counts in the exact hypergeometric sum; the correct exact sum stops at $K=5$.

A Small Sampling Fraction: A Useful but Not Exact Approximation

When $n/N$ is small, the correction factor for the variance is often close to 1. This supports the intuition that a binomial approximation may be useful, but it does not turn it into an exact model. The next example compares probabilities directly in a setting with a small sampling fraction.

Worked Example: Comparing “At Least One” Probabilities

A hypothetical quality check samples $n=3$ units without replacement from a batch of $N=100$, of which $K=20$ are marked. Find the exact and binomial-approximation probabilities of finding at least one marked unit. Also calculate the variance correction factor.

Method and model. The population fraction is $p=20/100=0.2$, and the sample fraction is $3/100=0.03$. For the exact probability, use the complement event of drawing no marked units. For the approximation, use $X_B\sim\operatorname{Binomial}(3,0.2)$.

Exact probability. There are $\binom{80}{3}=82160$ ways to choose three unmarked units, out of $\binom{100}{3}=161700$ total samples. Hence,

P(X\geq1) =1-\frac{\binom{80}{3}}{\binom{100}{3}} =1-\frac{82160}{161700} =\frac{79540}{161700} \approx 0.4919.

Binomial approximation. The chance of no marked units in three independent trials is $(1-0.2)^3$, so

P(X_B\geq1)=1-(0.8)^3=1-0.512=0.4880.

The difference is approximately $0.4918986-0.4880=0.0038986$, or $0.0039$ to four decimal places. The variance ratio is

\frac{N-n}{N-1}=\frac{100-3}{100-1} =\frac{97}{99}\approx0.9798.

The small sampling fraction and variance ratio near 1 are consistent with the modest difference observed here, but they do not prove that all event probabilities will have the same level of agreement. The calculation avoids the trap of calling the binomial probability exact simply because the sample is a small fraction of the population.

Choosing the Model in an Exam Problem

Start from the sampling mechanism in the problem statement. If the population has a fixed number of successes and the sample is drawn without replacement, the hypergeometric model is exact. If the problem explicitly states independent trials with constant success probability, use the binomial model. When a binomial approximation is invited or justified, use $p=K/N$, then assess whether it is suitable for the event being asked about.

Identify the population and sampling rule.
Record $N$, $K$, and $n$. Confirm whether selected items are returned or whether trials are explicitly independent.

Set the exact support.
For sampling without replacement, use the lower bound $\max(0,n-(N-K))$ and upper bound $\min(n,K)$. Do not sum over counts outside this range.

Assess the approximation.
Use $p=K/N$, inspect the fraction $n/N$, and check whether the binomial support includes impossible counts relevant to the requested event.

Calculate and interpret.
Use the exact formula when required or when the approximation is questionable. If using the binomial approximation, label it as approximate and compare the result with exact calculation when practical.

A frequent error is to treat the variance correction as a universal accuracy guarantee. It describes the variance ratio, not the error in an arbitrary event probability. Another is to claim that binomial support always includes impossible outcomes: the supports can coincide, such as when $n\leq K$ and $n\leq N-K$. Make the support comparison for the actual values in the problem.

Key takeaway: The hypergeometric model is exact for a fixed finite population sampled without replacement. The binomial model replaces dependent draws with independent trials of success probability $K/N$. A small sampling fraction can make that replacement useful, but check both the event probability and the feasible count range rather than relying on a rule of thumb alone.

Check Your Understanding

For each problem, identify the relevant count model and use exact arithmetic where requested. Probabilities are rounded to four decimal places.

A sample of 4 items is drawn without replacement from $N=60$ items, of which $K=12$ are marked. Find the exact probability of exactly one marked item and its binomial approximation. (Answer: exact $0.4256$; binomial $0.4096$)
A sample of $n=4$ is drawn without replacement from $N=12$, with $K=3$ successes. Find the exact probability of at least one success and the binomial approximation using $p=K/N$. (Answer: exact $369/495\approx0.7455$; binomial $1-(0.75)^4=0.6836$)
For $N=50$, $K=10$, and $n=5$, calculate the binomial variance and the hypergeometric variance. (Answer: binomial variance $0.8000$; hypergeometric variance $36/49\approx0.7347$)
A sample of $n=7$ is drawn without replacement from $N=18$, containing $K=4$ successes. State the exact and binomial supports. Why is the binomial approximation unsuitable for finding the exact probability of at least five successes? (Answer: exact support $0,1,2,3,4$; binomial support $0,1,\ldots,7$; the exact probability is 0, while the binomial model assigns positive probability to counts 5 through 7.)
For $N=200$, $K=40$, and $n=8$, calculate both variances and their ratio, then interpret the ratio. (Answer: binomial variance $1.2800$; hypergeometric variance $1.28(192/199)\approx1.2360$; ratio $192/199\approx0.9648$, so the hypergeometric variance is about 96.48% of the binomial variance.)

What You'll Learn