Comparing Hypergeometric and Binomial Probabilities

Why Compare the Two Models?

A binomial model and a hypergeometric model can describe counts of successes in samples of the same size, with the same success rate at the start. Their key difference is how the observations are generated. In a binomial model, trials are independent and each has the same success probability. In a hypergeometric model, a sample is selected without replacement from a finite population, so one draw changes the success proportion remaining for the next draw.

That distinction affects more than the variance. It can change the probability of a particular count or event, and it can make some counts impossible. The sampling fraction—the sample size as a proportion of the population—helps indicate how substantial these differences may be. The calculations below are aligned with the SOA VEE Mathematical Statistics syllabus.

Definition: Let a finite population contain $N$ items, of which $K$ are successes, and define the initial success rate $p=K/N$. If $n$ items are selected by simple random sampling without replacement, the number of successes $X$ has a hypergeometric distribution: $$ P(X=x)=\frac{\binom{K}{x}\binom{N-K}{n-x}}{\binom{N}{n}}. $$ If instead $n$ independent trials each have success probability $p$, then $X$ has a binomial distribution: $$ P(X=x)=\binom{n}{x}p^x(1-p)^{n-x}. $$

For the hypergeometric model, the possible values are the integers from $\max(0,n-(N-K))$ through $\min(n,K)$. The lower limit reflects the fact that the sample cannot contain more failures than the population has; the upper limit reflects the total number of successes available. A binomial variable with parameters $n$ and $p$, when $0<p<1$, can take any integer value from 0 through $n$.

When comparing the models, use the same $n$ and $p=K/N$. This makes the mean $np$ match. It does not make the distributions identical: independence holds in the binomial model, but not for sampling without replacement.

Worked Probability Comparisons

Worked Example: Exactly Two Successes in a Small Sample

A hypothetical inspection lot contains $N=20$ components, $K=8$ of which have a specified flaw. Four components are selected without replacement. Compare the probability of exactly two flawed components under the hypergeometric model with the probability under a binomial model using the same initial success rate and sample size. Also compare the variances.

Method and setup. The initial success rate is $p=K/N=8/20=0.4$, and the sample size is $n=4$. For the actual sample without replacement, use the hypergeometric probability: $$ P_{\mathrm{hyp}}(X=2) =\frac{\binom{8}{2}\binom{12}{2}}{\binom{20}{4}} =\frac{28\cdot66}{4845} =\frac{1848}{4845} =\frac{616}{1615} \approx 0.3814. $$

For the binomial model, the success probability stays at $0.4$ on each independent trial: $$ P_{\mathrm{bin}}(X=2) =\binom{4}{2}(0.4)^2(0.6)^2 =6(0.16)(0.36) =0.3456. $$ Thus, exactly two successes are more likely under the hypergeometric model in this example.

Both means equal $np=4(0.4)=1.6$. The binomial variance is $np(1-p)=4(0.4)(0.6)=0.96$. The hypergeometric variance is $$ np(1-p)\frac{N-n}{N-1} =0.96\frac{16}{19} =\frac{15.36}{19} \approx 0.8084. $$ The sampling fraction is $n/N=4/20=0.20$, so the finite-population adjustment is material. Check: both models have mean $1.6$, while the hypergeometric variance is smaller and its probability of exactly two is $0.3814$, not $0.3456$. The common mistake to avoid is treating the without-replacement sample as independent just because its initial success rate is $0.4$.

Worked Example: Comparing the Probability of No Successes

A hypothetical batch has $N=12$ units, including $K=3$ units with a particular feature. A sample of $n=3$ is drawn without replacement. Compare the probability of no units with the feature under the hypergeometric model and under the binomial model with the same $p$ and $n$. Then compare the variances.

Method and setup. Here $p=3/12=0.25$. Under sampling without replacement, getting no successes means choosing all three units from the nine failures: $$ P_{\mathrm{hyp}}(X=0) =\frac{\binom{3}{0}\binom{9}{3}}{\binom{12}{3}} =\frac{1\cdot84}{220} =\frac{21}{55} \approx 0.3818. $$

Under the binomial model, the probability of failure on each independent trial is $1-p=0.75$, so $$ P_{\mathrm{bin}}(X=0) =\binom{3}{0}(0.25)^0(0.75)^3 =0.421875 \approx 0.4219. $$ In this case, no successes are less likely under the hypergeometric model.

The common mean is $np=3(0.25)=0.75$. The binomial variance is $$ 3(0.25)(0.75)=0.5625. $$ The hypergeometric variance is $$ 0.5625\frac{12-3}{12-1} =0.5625\frac{9}{11} \approx 0.4602. $$ The sampling fraction is $3/12=0.25$. Check: the probability calculations use a denominator of $\binom{12}{3}=220$ for the unordered samples and agree with the variance comparison: the without-replacement count is less variable. The mistake to avoid is concluding that the hypergeometric probability must be higher just because the hypergeometric variance is lower; the direction depends on the event.

Worked Example: A Large Sampling Fraction

A hypothetical collection contains $N=10$ records, $K=4$ of which meet a review criterion. A reviewer samples $n=5$ records without replacement. Compare the probability of finding no qualifying records under the two models, and assess the effect of the sampling fraction on variance.

Method and setup. The common success rate is $p=4/10=0.4$. The hypergeometric probability of zero successes is $$ P_{\mathrm{hyp}}(X=0) =\frac{\binom{4}{0}\binom{6}{5}}{\binom{10}{5}} =\frac{1\cdot6}{252} =\frac{1}{42} \approx 0.0238. $$

The binomial probability with $n=5$ and $p=0.4$ is $$ P_{\mathrm{bin}}(X=0) =(1-0.4)^5 =(0.6)^5 =0.07776 \approx 0.0778. $$ The hypergeometric probability is much smaller. Once five of ten records have been selected, it is difficult to avoid all four qualifying records.

The common mean is $np=5(0.4)=2$, and the binomial variance is $5(0.4)(0.6)=1.2$. Applying the finite-population correction gives $$ \operatorname{Var}_{\mathrm{hyp}}(X) =1.2\frac{10-5}{10-1} =1.2\frac{5}{9} =\frac{2}{3} \approx 0.6667. $$ The sampling fraction is $5/10=0.50$, which is large. Check: the hypergeometric support is $0$ through $4$, because the population contains only four successes; the binomial model also permits $X=5$. The common mistake is to ignore this finite-population limit and assume the models remain close when half the population is sampled.

Why the Variances Differ

The mean of the number of successes is $np$ under either model. The variance differs because distinct draws in a sample without replacement are negatively associated: after one success is selected, fewer successes remain available for later draws.

For a derivation, write the hypergeometric count as $X=I_1+\cdots+I_n$, where $I_j=1$ if the $j$th draw is a success and $I_j=0$ otherwise. Each draw position has $P(I_j=1)=p$, so $E[I_j]=p$ and $\operatorname{Var}(I_j)=p(1-p)$. For distinct positions $j$ and $k$, the probability that both are successes is $$ P(I_j=1,I_k=1)=\frac{K}{N}\frac{K-1}{N-1}. $$ Therefore, $$ \operatorname{Cov}(I_j,I_k) =\frac{K}{N}\frac{K-1}{N-1}-p^2 =-\frac{p(1-p)}{N-1}. $$

Adding the variances and covariances of the indicators gives $$ \begin{aligned} \operatorname{Var}(X) &=n p(1-p)+n(n-1)\left(-\frac{p(1-p)}{N-1}\right)\\ &=np(1-p)\left(1-\frac{n-1}{N-1}\right)\\ &=np(1-p)\frac{N-n}{N-1}. \end{aligned} $$

The binomial variance is $np(1-p)$, because its trial indicators are independent and therefore have covariance zero. The factor $(N-n)/(N-1)$ is the finite-population correction to the variance. It equals 1 when $n=1$, is less than 1 when $1<n\leq N$, and equals 0 when $n=N$. If the whole population is observed, the total number of successes is fixed at $K$, so zero variance is appropriate.

Interpreting Probability Differences

A smaller hypergeometric variance does not imply that every particular hypergeometric probability is smaller. The distribution concentrates more tightly around its mean, but probabilities in different parts of the range can move in opposite directions. The examples show this: the hypergeometric probability of exactly two successes exceeded the binomial probability in one case, while the probability of zero successes was lower in the other cases.

The sampling fraction $n/N$ is a useful guide, but it is not by itself a complete probability calculation. When the fraction is small, successive draws change the remaining success rate relatively little, and the two models often give similar probabilities. As the fraction grows, the finite-population constraint matters more, both through the variance and through the possible counts. Always calculate the requested event using the correct model rather than applying a universal rule about which model assigns the larger probability.

Common exam trap: Using $p=K/N$ correctly does not make a hypergeometric sample binomial. For the binomial model, trials must be independent with a constant success probability. For a finite sample drawn without replacement, use the hypergeometric probability formula. Compare event probabilities directly; use the finite-population correction to compare variances, not to determine the direction of every probability difference.

Identify the sampling process.
Without replacement from a finite population calls for a hypergeometric model; independent trials with constant success probability call for a binomial model.

Match the parameters.
For a fair comparison, set $p=K/N$ and use the same sample size $n$ in both models.

Check support and calculate the event.
Confirm that the requested count is possible, then use the probability formula for each model.

Compare variability and context.
Both means are $np$; the hypergeometric variance is multiplied by $(N-n)/(N-1)$. Interpret that factor alongside the sampling fraction.

Check Your Understanding

For each problem, compare the stated event under both models using the same initial success rate and sample size. Round probabilities to four decimal places and variances to four decimal places.

A population has $N=15$ items, $K=6$ successes, and a sample size $n=2$. Compare the probability of two successes and the two variances. (Answer: $P_{\mathrm{hyp}}(X=2)=0.1429$, $P_{\mathrm{bin}}(X=2)=0.1600$; variances $0.4457$ and $0.4800$, respectively.)
A population has $N=16$ items, $K=4$ successes, and $n=4$. Compare the probability of zero successes and the variances. (Answer: $P_{\mathrm{hyp}}(X=0)=495/1820\approx0.2720$, $P_{\mathrm{bin}}(X=0)=0.75^4\approx0.3164$; variances $0.6000$ and $0.7500$, respectively.)
A population has $N=20$ items, $K=5$ successes, and $n=4$. Compare the probability of exactly one success and the variances. (Answer: $P_{\mathrm{hyp}}(X=1)=2275/4845\approx0.4696$, $P_{\mathrm{bin}}(X=1)=0.4219$; variances $0.6316$ and $0.7500$, respectively.)
A population has $N=8$ items, $K=2$ successes, and $n=4$. Compare the probability of at least one success and the variances. (Answer: $P_{\mathrm{hyp}}(X\geq1)=11/14\approx0.7857$, $P_{\mathrm{bin}}(X\geq1)=1-0.75^4\approx0.6836$; variances $0.4286$ and $0.7500$, respectively.)

What You'll Learn