Hypergeometric Probabilities Over Count Ranges

Why Count Ranges Need Careful Translation

A question about sampling without replacement may ask for the probability of observing a count within a range, rather than one exact count. The calculation uses the same hypergeometric probability mass function as an exact-count problem, but first the range wording must be translated into the correct integer outcomes. In particular, “at least,” “less than,” and a noninteger endpoint can determine whether a boundary value is included.

Suppose a population has $N$ items, of which $K$ are classified as successes. A simple random sample of $n$ items is selected without replacement, and $X$ denotes the number of successes in the sample. The number of successes must be an integer. Its feasible values are restricted by the population and sample sizes.

Definition: Under the hypergeometric model, each sample of size $n$ from the $N$ distinct population items is equally likely, and $X$ counts how many of the $K$ successes it contains. Its probability mass function is $$ P(X=x)=\frac{\binom{K}{x}\binom{N-K}{n-x}}{\binom{N}{n}}, $$ for integers $x$ satisfying $\max(0,n-(N-K))\le x\le \min(n,K)$. For any other integer $x$, $P(X=x)=0$.

The assumptions matter: the population has a fixed number $K$ of successes, the sample is selected without replacement, and every size-$n$ subset is equally likely. A binomial calculation instead models independent trials with a constant success probability, so it is not the direct model for a fixed-size sample without replacement.

Translate the Range Before Calculating

Write down the integer outcomes permitted by the event, then intersect them with the support of $X$. The support restriction can remove outcomes even when the range wording allows them. For instance, a sample cannot contain more successes than either the sample size $n$ or the population total $K$.

Key rule: A range probability is the sum of the probability mass at every feasible integer count included in the event. Inclusive endpoints are included; strict endpoints are excluded. Noninteger boundaries are not rounded: test the integers against the original inequalities.

For integer bounds $a$ and $b$, the event $a\le X\le b$ includes the integers from $a$ through $b$, while $a<X<b$ includes only the integers strictly between them. For a noninteger bound such as $1.5<X\le4$, the qualifying integers are $2,3,4$. After translating, discard any count outside the hypergeometric support.

Translate the inequalities.
List the integer values satisfying the event, respecting every strict or inclusive endpoint.

Apply the support.
Keep only counts from $\max(0,n-(N-K))$ through $\min(n,K)$.

Calculate and add.
Evaluate the hypergeometric mass at each remaining count and sum the terms.

Worked Range Probabilities

Worked Example: An Inclusive Range of Counts

A sealed tray contains $N=12$ components, exactly $K=5$ of which have a specified feature. A quality analyst selects $n=4$ components uniformly at random without replacement. Find the probability that the sample contains at least 1 and at most 3 components with the feature.

Method. This is a hypergeometric count. The requested event is $1\le X\le3$, so its possible counts are $1,2,3$. The support is $0\le X\le4$, and all three requested counts are feasible.

The total number of samples is $\binom{12}{4}=495$. The numbers of samples for the three counts are \[ \binom{5}{1}\binom{7}{3}=5(35)=175,\qquad \binom{5}{2}\binom{7}{2}=10(21)=210, \] and \[ \binom{5}{3}\binom{7}{1}=10(7)=70. \] Thus the range probability is

P(1\le X\le3) =\frac{175+210+70}{495} =\frac{455}{495} =\frac{91}{99} \approx 0.9192.

Check. The favorable sample count is $455$, which is no greater than the $495$ possible samples. The reported probability is rounded to four decimal places. Common mistake avoided: treating “at least 1 and at most 3” as a single exact count would omit two of the three included masses.

Worked Example: Strict Endpoints

A collection has $N=10$ devices, $K=4$ of which are flagged for review. Five devices are selected uniformly at random without replacement. Calculate the probability that the number flagged is strictly greater than 2 and strictly less than 5.

Method. The event is $2<X<5$. Since $X$ is an integer, only $X=3$ and $X=4$ qualify. The support is $0\le X\le4$; both counts are feasible. There are $\binom{10}{5}=252$ possible samples.

For $X=3$, choose 3 of the 4 flagged devices and 2 of the 6 unflagged devices: $\binom{4}{3}\binom{6}{2}=4(15)=60$. For $X=4$, choose all 4 flagged devices and 1 of the 6 unflagged devices: $\binom{4}{4}\binom{6}{1}=1(6)=6$. Therefore,

P(2<X<5) =\frac{60+6}{252} =\frac{66}{252} =\frac{11}{42} \approx 0.2619.

Check. The two qualifying counts account for $66$ of the $252$ possible samples. Common mistake and correction: including $X=2$ would produce the incorrect numerator $120+60+6=186$, because $\binom{4}{2}\binom{6}{3}=120$. But $2$ is excluded by $2<X$; the correct numerator is $60+6=66$. The probability is rounded to four decimal places.

Worked Example: Noninteger Bounds and Feasible Counts

A shipment contains $N=14$ sensors, $K=5$ of which have a certain defect. An inspector samples $n=6$ sensors uniformly at random without replacement. Find the probability that the defect count $X$ satisfies $1.2<X\le4$.

Method. Test integer counts against the stated inequalities rather than rounding either boundary. The counts satisfying $1.2<X\le4$ are $2,3,4$. The support is $0\le X\le5$, so all three are feasible. The total number of samples is $\binom{14}{6}=3003$.

For $X=2$, the number of samples is $\binom{5}{2}\binom{9}{4}=10(126)=1260$. For $X=3$, it is $\binom{5}{3}\binom{9}{3}=10(84)=840$. For $X=4$, it is $\binom{5}{4}\binom{9}{2}=5(36)=180$. Adding the disjoint cases gives

P(1.2<X\le4) =\frac{1260+840+180}{3003} =\frac{2280}{3003} =\frac{760}{1001} \approx 0.7592.

Check. The integer $1$ fails the strict lower bound, while $4$ satisfies the inclusive upper bound. The numerator $2280$ is less than the denominator $3003$. The probability is rounded to four decimal places. Common mistake avoided: rounding $1.2$ to $1$ before applying the strict inequality could wrongly include $X=1$.

Why the Range Formula Is a Sum

Once the qualifying integer values have been identified, the range calculation follows from additivity of probability. The events $X=x$ for different integers $x$ cannot occur together: a sample cannot have two different success counts. They are therefore disjoint, and their union is exactly the requested range event.

Let $L$ be the set of feasible integer values satisfying the range inequalities. By the addition rule for disjoint events,

P(X\in L) =P\left(\bigcup_{x\in L}\{X=x\}\right) =\sum_{x\in L}P(X=x) =\sum_{x\in L} \frac{\binom{K}{x}\binom{N-K}{n-x}}{\binom{N}{n}}.

The last equality substitutes the hypergeometric probability mass function for each $x$. The set $L$ must include only feasible counts; equivalently, any infeasible count contributes probability zero. This reasoning explains why every qualifying mass is included exactly once and why masses outside the event must not be added.

Exam Checks and Common Traps

Range questions are often decided by the translation step rather than by difficult arithmetic. Before evaluating combinations, record the event in symbols and list the integer counts. Then verify that each term uses $x$ successes and $n-x$ non-successes, and that the terms share the same denominator $\binom{N}{n}$.

Confusing “at least” with “greater than.” “At least 2” means $X\ge2$, including 2. “Greater than 2” excludes 2.
Rounding a boundary instead of testing integers. For $1.2<X\le4$, test each integer against both inequalities; the valid counts are $2,3,4$.
Ignoring the support. A count can satisfy the written range but still be impossible given the sample and population composition.
Using a binomial mass without its assumptions. Sampling a fixed number of items without replacement from a finite population gives a hypergeometric model under the stated uniform sampling design.
Adding a boundary mass twice. Write one list of distinct qualifying counts before summing.

Key takeaway: Translate the bounds into integers, intersect those integers with the hypergeometric support, and sum the exact-count probabilities for the outcomes that remain.

Check Your Understanding

For each problem, identify the qualifying integer counts before calculating. Each sample is selected uniformly at random without replacement.

A population has $N=9$ items, $K=3$ successes, and a sample of $n=2$. Find $P(0\le X\le1)$. (Answer: $11/12\approx0.9167$)
A population has $N=11$ items, $K=4$ successes, and a sample of $n=3$. Find $P(1<X\le3)$. (Answer: $46/165\approx0.2788$)
A population has $N=13$ items, $K=5$ successes, and a sample of $n=4$. Find $P(0.5<X<3.5)$. (Answer: $128/143\approx0.8951$)
A population has $N=12$ items, $K=2$ successes, and a sample of $n=5$. Find $P(2\le X\le4)$, taking account of the support. (Answer: $5/33\approx0.1515$)
A population has $N=16$ items, $K=7$ successes, and a sample of $n=5$. Find $P(1\le X<4)$. (Answer: $93/104\approx0.8942$)

What You'll Learn