The Support of a Hypergeometric Count

Why the Support Matters

A hypergeometric count records how many successes appear in a sample drawn without replacement from a finite population. Before calculating any probability for such a count, determine which values it can actually take. A count outside that set is impossible; a count inside it is feasible under the standard sampling model. Knowing the support also prevents errors when setting up later probability calculations.

For this tutorial, a success means the category being counted, not necessarily a desirable outcome. The same reasoning applies whether the category is defective parts, eligible records, or a particular type of item. The results are aligned with the SOA VEE Mathematical Statistics syllabus.

Definition: Suppose a population has $N$ objects, of which $K$ are successes and $N-K$ are failures. Select $n$ objects by simple random sampling without replacement, where $N$, $K$, and $n$ are integers satisfying $0\le K\le N$ and $0\le n\le N$. Let $X$ be the number of successes selected. The support of $X$ is the set of values that $X$ can take with positive probability under this sampling model.

The model assumes every sample of size $n$ is possible under the sampling design: each distinct subset of that size has positive probability of being selected. The support is determined by the population composition and sample size, not by the order in which selected objects are listed. This tutorial concerns possible counts only; it does not calculate the probabilities attached to those counts.

Deriving the Smallest and Largest Counts

Let $x$ be a proposed number of successes in the sample. Then the sample must contain $n-x$ failures. A feasible count must satisfy all of the following: it cannot be negative; it cannot exceed the total number of successes; it cannot exceed the sample size; and the sample cannot contain more failures than the population has available.

0\le x,\qquad x\le K,\qquad x\le n,\qquad 0\le n-x\le N-K.

The last pair of inequalities is the constraint on the selected failures. Its upper bound gives $n-x\le N-K$, which is equivalent to $x\ge n-(N-K)$. Combining the lower bounds and combining the upper bounds gives the complete candidate range.

Theorem: Under simple random sampling without replacement, the support of the hypergeometric count $X$ is $$ \left\{x\in\mathbb{Z}:\ \max\bigl(0,n-(N-K)\bigr)\le x\le \min(n,K)\right\}. $$ Equivalently, the smallest possible count is $x_{\min}=\max(0,n-(N-K))$, and the largest possible count is $x_{\max}=\min(n,K)$. Every integer between these bounds is possible.

The upper bound reflects two separate limits: the sample cannot contain more successes than the $K$ in the population, or more successes than the $n$ objects selected. The lower bound reflects a different constraint. If the population has too few failures to fill the whole sample, some successes must be selected. When there are enough failures to fill the sample, zero successes remains possible.

Why the Bounds Give the Whole Support

The inequalities above show that any feasible $x$ must lie between the stated bounds. To confirm that these bounds do not leave gaps, take any integer $x$ in that range. Because $x\ge0$ and $x\le K$, there are enough population successes to choose $x$ of them. Because $x\le n$, the required number of failures, $n-x$, is nonnegative. Finally, $x\ge n-(N-K)$ implies $n-x\le N-K$, so there are enough failures to choose the rest of the sample.

Thus it is possible to choose $x$ successes and $n-x$ failures. Together those choices make a sample of size $n$, and the count in that sample is $x$. Every integer in the bounds is therefore feasible. Under the stated sampling model, each such choice of objects has positive probability, so each of those counts belongs to the support. No count outside the bounds can occur because it violates at least one of the population or sample-size constraints.

Record the population composition.
Write the number of successes as $K$ and calculate the number of failures as $N-K$.

Apply both constraints.
The maximum is limited by the available successes and the sample size. The minimum is determined by how many successes are forced when the available failures cannot fill the sample.

State the integer range.
Include both endpoints and every integer between them. Check that the proposed range is consistent with the sample size and population counts.

Worked Examples

Worked Example: A Sample That Could Contain No Successes

A hypothetical archive contains 48 records. Thirteen are flagged for a particular review category, and the remaining records are not flagged. A simple random sample of 16 records is selected without replacement. Determine the support of the number $X$ of flagged records in the sample.

Method. Identify the population size, success count, failure count, and sample size, then apply the support bounds. Here $N=48$, $K=13$, and $n=16$. The number of failures is $N-K=48-13=35$.

The smallest possible count is \[ x_{\min}=\max(0,n-(N-K))=\max(0,16-35)=\max(0,-19)=0. \] The largest possible count is \[ x_{\max}=\min(n,K)=\min(16,13)=13. \]

The support is therefore $0,1,2,\ldots,13$. The endpoints are feasible: a sample with zero flagged records can be filled using 16 of the 35 unflagged records, and a sample with 13 flagged records can include all 13 flagged records plus 3 of the 35 unflagged records. The endpoints and all intermediate counts satisfy the available-category constraints. The common mistake to avoid is treating the upper endpoint as 16 just because the sample has 16 records; there are only 13 flagged records in the population.

Worked Example: A Sample That Must Contain Successes

A hypothetical shipment has 30 components, including 22 of a specified type and 8 of other types. An inspector selects 25 components without replacement. Determine the support of $X$, the number of components of the specified type in the sample.

Method. Use the failure count to determine whether the sample can be filled without selecting any successes. The parameters are $N=30$, $K=22$, $N-K=30-22=8$, and $n=25$.

At most 8 failures can be selected. Therefore at least $25-8=17$ selected components must be successes: \[ x_{\min}=\max(0,25-8)=17. \] The sample cannot contain more successes than either the sample size or the number of successes in the population: \[ x_{\max}=\min(25,22)=22. \]

Thus the support is $17,18,19,20,21,22$. For the lower endpoint, select all 8 failures and 17 of the 22 successes, giving $8+17=25$ objects. For the upper endpoint, select all 22 successes and 3 failures, again giving $22+3=25$ objects. Both are feasible. A common incorrect calculation would give a lower bound of zero and the range $0,\ldots,22$. That ignores that 25 objects must be selected even though the entire population contains only 8 failures; the corrected lower bound is 17.

Worked Example: Checking Proposed Counts Against the Support

A hypothetical review group contains 73 files, of which 28 meet a specified criterion and 45 do not. A random sample of 52 files is selected without replacement. Determine the support of $X$, then decide whether counts of 6, 7, 17, and 29 are feasible.

Method. First find the support; then compare each proposed count with both endpoints. Here $N=73$, $K=28$, the number of failures is $73-28=45$, and $n=52$.

Since there are 45 failures but 52 files must be selected, at least $52-45=7$ successes are required: \[ x_{\min}=\max(0,52-45)=7. \] The upper bound is \[ x_{\max}=\min(52,28)=28. \] Therefore the support is every integer from 7 through 28, inclusive.

A count of 6 is not feasible because it is below the minimum. A count of 7 is feasible: it uses 7 successes and $52-7=45$ failures, exactly the number available. A count of 17 is feasible: it uses 17 successes and $52-17=35$ failures, both within the available totals. A count of 29 is not feasible because it exceeds the 28 successes in the population. The final check is that each feasible count uses 52 files in total and does not request more of either category than exists. The trap is checking only whether a count is less than the total number of successes; the sample size also forces the lower bound.

Common Mistakes and Checks

The most frequent error is to set the lower bound to zero automatically. Zero is possible only if there are at least $n$ failures available, or $N-K\ge n$. If this condition fails, the sample must include successes. The reverse error can also occur: a positive lower bound should not be imposed merely because successes are common in the population. Compare the sample size directly with the number of failures.

Other useful checks are straightforward:

The upper endpoint cannot exceed $K$, the number of successes in the population, or $n$, the sample size.
The lower endpoint cannot be negative and must be large enough that the remaining sample positions can be filled with failures.
The bounds are inclusive. If an endpoint satisfies the constraints, it is part of the support.
Values in the support are integers. A noninteger count is not a possible count of objects.
Finding the support does not assign probabilities to its values. A feasible count is not necessarily an equally likely count.

Key takeaway: For $N$ objects containing $K$ successes and $N-K$ failures, a sample of size $n$ has success-count support from $\max(0,n-(N-K))$ through $\min(n,K)$, including every integer in between.

Check Your Understanding

For each problem, identify the success and failure totals, calculate both endpoints, and state the requested feasibility result.

A hypothetical collection has 20 items, 6 of which are marked for inspection. A sample of 9 is selected without replacement. State the support for the number of marked items. (Answer: $0,1,2,3,4,5,6$)
A set of 44 records has 31 records in a target category and 13 outside it. A sample of 20 is drawn. Find the smallest and largest possible target-category counts. (Answer: minimum $7$, maximum $20$; support $7,8,\ldots,20$)
A hypothetical population contains 57 devices, 14 of which have a specified feature. If 49 devices are selected, determine the support of the feature count. (Answer: $6,7,\ldots,14$)
A group of 36 cases contains 9 successes and 27 failures. A sample of 28 is taken. Determine the support and decide whether counts 0, 1, 9, and 10 are feasible. (Answer: support $1,2,\ldots,9$; 0 and 10 are infeasible, while 1 and 9 are feasible)
A population has 90 objects, 37 successes, and 53 failures. A sample of 61 is drawn. Determine the support and assess whether 7, 8, 37, and 38 successes are possible. (Answer: support $8,9,\ldots,37$; 7 and 38 are impossible, while 8 and 37 are possible)

What You'll Learn