Symmetries of the Hypergeometric Distribution

Why Hypergeometric Symmetries Help

A hypergeometric probability can sometimes be easier to calculate after changing what is counted. If a sample contains nearly all the population, it may be simpler to count the few omitted items. If the event describes many successes, counting the comparatively few failures may require fewer terms. These are not approximations: each change gives an exactly equivalent description of the same random sample.

The transformations are useful only when the event is translated carefully. Changing the label “success” to “failure” changes the population count and the observed count; viewing a sample through its omitted items changes the sample size and relates the two counts by subtraction. This tutorial develops those identities and shows how to use them. The methods are aligned with the SOA VEE Mathematical Statistics syllabus.

Definition: A population has $N$ distinct items, of which $K$ are designated successes and $N-K$ are failures. A simple random sample of size $n$ is selected without replacement, with $0\leq n\leq N$. Let $X$ be the number of successes in the sample. Then $X\sim\operatorname{Hypergeometric}(N,K,n)$, with $$ P(X=x)=\frac{\binom{K}{x}\binom{N-K}{n-x}}{\binom{N}{n}} $$ for feasible integer $x$. The denominator counts all samples of size $n$, and the numerator counts those with $x$ successes and $n-x$ failures.

Two Basic Ways to Re-express the Count

First, count the failures in the sample instead of the successes. There are $N-K$ failures in the population, and the sample still has size $n$. If $F=n-X$ is the number of sampled failures, then

F\sim\operatorname{Hypergeometric}(N,N-K,n),\qquad X=n-F.

Second, consider the $N-n$ items that were not selected. Call their number of successes $Z$. Since all $K$ population successes are either selected or omitted,

Z\sim\operatorname{Hypergeometric}(N,K,N-n),\qquad X=K-Z.

The first transformation keeps the sample size fixed and switches the success label. The second keeps the success label fixed and switches from the selected items to the omitted items. Both retain sampling without replacement from the same finite population.

Theorem: For any feasible integer $x$, the hypergeometric probability mass has the equivalent forms $$ \frac{\binom{K}{x}\binom{N-K}{n-x}}{\binom{N}{n}} = \frac{\binom{N-K}{n-x}\binom{K}{x}}{\binom{N}{n}} = \frac{\binom{K}{K-x}\binom{N-K}{N-n-(K-x)}}{\binom{N}{N-n}}. $$ The first equality expresses the same sample by counting failures. The last expresses it by the successes and failures among the omitted items.

Why the Identities Hold

Each possible sample of size $n$ is equally likely. A sample with $x$ successes contains $n-x$ failures. Relabeling success and failure therefore describes the same sample by choosing $n-x$ items from the $N-K$ failures and $x$ items from the $K$ successes. This gives the identity

P(X=x)=P(F=n-x) =\frac{\binom{N-K}{n-x}\binom{K}{x}}{\binom{N}{n}}.

For the omitted-items identity, a sample with $x$ selected successes leaves $K-x$ successes among the $N-n$ omitted items. It also leaves $N-n-(K-x)$ failures omitted. Choosing those omitted items has the same number of possibilities as choosing the original sample: a subset and its complement determine one another uniquely. In particular,

\binom{N}{N-n}=\binom{N}{n},\qquad \binom{K}{K-x}=\binom{K}{x},\qquad \binom{N-K}{N-n-K+x}=\binom{N-K}{n-x}.

Substituting these equalities into the probability formula proves that the probability for $X=x$ equals the probability for $Z=K-x$ when the omitted items are treated as the sample. These are counting identities, not distributional approximations.

Transforming Events and Their Bounds

A transformation is useful only if the event is transformed along with the random variable. The main translations are:

Original count	Equivalent count	Event translation
Selected successes $X$	Selected failures $F=n-X$	$X\geq a$ becomes $F\leq n-a$
Selected successes $X$	Omitted successes $Z=K-X$	$X\geq a$ becomes $Z\leq K-a$
Selected successes $X$	Omitted failures $W=N-n-K+X$	$X\leq b$ becomes $W\leq N-n-K+b$

The third row combines both changes of viewpoint: $W$ counts failures among the omitted items, so $W\sim\operatorname{Hypergeometric}(N,N-K,N-n)$. It follows because the omitted set has $N-n$ items, of which $N-n-(K-X)$ are failures.

Name the original model.
Record $N$, $K$, $n$, the variable being counted, and the event of interest.

Choose the smaller or clearer count.
Use $n-X$ to count sampled failures, or $K-X$ to count omitted successes. If useful, count omitted failures instead.

Translate the event before calculating.
Apply the relevant subtraction to every endpoint, preserving whether the inequality reverses.

Check the transformed model and answer.
Its sample size must be $n$ for sampled failures and $N-n$ for omitted items. Confirm the transformed event is feasible and, when practical, verify by counting in the original model.

Worked Example: Counting Failures for a High-Success Event

A hypothetical archive contains $N=22$ sealed packets, including $K=17$ packets with a designated feature and $N-K=5$ without it. A simple random sample of $n=5$ packets is selected without replacement. Find the probability that at least four sampled packets have the feature.

Method. Let $X$ be the number with the feature. Directly, the event is $X\geq4$, so it includes $X=4$ and $X=5$. Instead let $F=5-X$ count sampled packets without the feature. Then $F\sim\operatorname{Hypergeometric}(22,5,5)$, and $X\geq4$ is equivalent to $F\leq1$.

There are $\binom{22}{5}=26{,}334$ equally likely samples. The probability of no failures is $\binom{5}{0}\binom{17}{5}=1(6{,}188)=6{,}188$. The probability of exactly one failure is $\binom{5}{1}\binom{17}{4}=5(2{,}380)=11{,}900$. Thus the favorable count is $6{,}188+11{,}900=18{,}088$, and

P(X\geq4)=P(F\leq1) =\frac{18{,}088}{26{,}334} \approx0.6868687 \approx\boxed{0.6869}.

The answer is rounded to four decimal places. As a check, direct counting gives $\binom{17}{4}\binom{5}{1}+\binom{17}{5}\binom{5}{0}=11{,}900+6{,}188=18{,}088$, the same favorable count. The method avoids summing in terms of the common feature when failures provide the more natural count.

Worked Example: Viewing a Large Sample Through Omitted Items

A hypothetical collection has $N=18$ components, of which $K=7$ have a specified characteristic and $11$ do not. A sample of $n=15$ is drawn without replacement. Find the probability that the sample contains at least six components with the characteristic.

Method. Only $N-n=3$ components are omitted. Let $Z$ be the number of characteristic components among those three; $Z\sim\operatorname{Hypergeometric}(18,7,3)$. Since $X=7-Z$, the event $X\geq6$ becomes $Z\leq1$. The sample space for the omitted set has $\binom{18}{3}=816$ equally likely subsets.

For $Z=0$, there are $\binom{7}{0}\binom{11}{3}=1(165)=165$ omitted sets. For $Z=1$, there are $\binom{7}{1}\binom{11}{2}=7(55)=385$. Therefore,

P(X\geq6)=P(Z\leq1) =\frac{165+385}{816} =\frac{550}{816} \approx\boxed{0.6740}.

The result is rounded to four decimal places. Check by counting the selected sample directly: $X=6$ contributes $\binom{7}{6}\binom{11}{9}=7(55)=385$, and $X=7$ contributes $\binom{7}{7}\binom{11}{8}=1(165)=165$. The same total of $550$ favorable samples confirms the transformation.

Worked Example: Combining Both Symmetries

A hypothetical population has $N=24$ items, with $K=9$ successes and $15$ failures. A sample of $n=20$ is selected without replacement. Find the probability that the sample has at most seven successes.

Method. Let $X$ count selected successes. The event is $X\leq7$. There are only $N-n=4$ omitted items, so count their failures instead. Let $W$ be the number of omitted failures. The omitted set has four items drawn from a population with $15$ failures and $9$ successes, so $W\sim\operatorname{Hypergeometric}(24,15,4)$. If $X=x$, then the omitted successes number $9-x$, and therefore $W=4-(9-x)=x-5$. Thus $X\leq7$ becomes $W\leq2$.

There are $\binom{24}{4}=10{,}626$ possible omitted sets. For $W=0$, the count is $\binom{15}{0}\binom{9}{4}=126$. For $W=1$, it is $\binom{15}{1}\binom{9}{3}=15(84)=1{,}260$. For $W=2$, it is $\binom{15}{2}\binom{9}{2}=105(36)=3{,}780$. The total is $126+1{,}260+3{,}780=5{,}166$, giving

P(X\leq7)=P(W\leq2) =\frac{5{,}166}{10{,}626} \approx\boxed{0.4862}.

The answer is rounded to four decimal places. To check, the original feasible values are $X=5,6,7$. Their counts are $126(1)=126$, $84(15)=1{,}260$, and $36(105)=3{,}780$, respectively, again totaling $5{,}166$. The combined transformation has changed both the counted label and the sample viewpoint while preserving the event probability.

Common Mistakes and Exam Traps

Changing the count but not its event. If $F=n-X$, then $X\geq a$ is $F\leq n-a$, not $F\geq n-a$. Subtraction reverses the direction of an inequality.
Using the wrong population count. When counting sampled failures, the population has $N-K$ successes under the new label, not $K$. When counting omitted successes, the sample size is $N-n$, not $n$.
Confusing omitted successes and omitted failures. The omitted successes are $K-X$; omitted failures are $N-n-K+X$. These are different counts and generally produce different event bounds.
Summing impossible values. A transformed variable must still obey its support. For instance, it cannot count fewer than zero items or more items than are available in its sample or population.

A deliberately incorrect setup for the second worked example would be to write $Z\sim\operatorname{Hypergeometric}(18,7,15)$, treating the omitted successes as if they were counted in a sample of size $15$. That is not the omitted set: only three items were omitted. The correction is $Z\sim\operatorname{Hypergeometric}(18,7,3)$, followed by $X=7-Z$. Always identify which collection is being sampled before substituting parameters.

Key takeaway: The same hypergeometric sample can be described by selected successes, selected failures, omitted successes, or omitted failures. The counts are related exactly by $F=n-X$, $Z=K-X$, and $W=N-n-K+X$. Choose the clearest count, transform the event bounds explicitly, and keep the corresponding sample size and population label consistent.

Check Your Understanding

For each problem, identify the transformed count and translate the event before evaluating the probability. Round probabilities to four decimal places.

A population has $N=16$ items, $K=12$ successes, and a sample of $n=4$ is drawn without replacement. Find the probability of at least three successes by counting sampled failures. (Answer: $0.7555$)
A population has $N=14$ items, $K=5$ successes, and $n=12$ items are selected. Find the probability of four or five successes by counting omitted successes. (Answer: $0.8901$)
A population has $N=25$ items, $K=10$ successes, and a sample of $n=22$ is selected. Find the probability of at most nine successes by counting omitted successes. (Answer: $0.8022$)
A population has $N=19$ items, $K=14$ successes, and $n=6$ are selected. Find the probability of exactly two successes by counting sampled failures. (Answer: $0.0168$)
A population has $N=24$ items, $K=9$ successes, and a sample of $n=20$ is selected. Find the probability of at most seven successes by counting omitted failures. (Answer: $0.4862$)

Original count	Equivalent count	Event translation
Selected successes \(X\)	Selected failures \(F=n-X\)	\(X\geq a\) becomes \(F\leq n-a\)
Selected successes \(X\)	Omitted successes \(Z=K-X\)	\(X\geq a\) becomes \(Z\leq K-a\)
Selected successes \(X\)	Omitted failures \(W=N-n-K+X\)	\(X\leq b\) becomes \(W\leq N-n-K+b\)

What You'll Learn