Why Finite-Population Sampling Needs Its Own Model
Suppose an insurer reviews a fixed number of files from a finite set of submitted claims. Each file either contains a specified feature or does not, and a file that has been reviewed is not returned to the set for selection. The number of selected files with the feature is a random count, but the draws are not independent: after one file is selected, the population available for the next selection has changed.
The hypergeometric model describes this type of count. Its defining ingredients are a finite population divided into two outcome categories, a fixed-size sample selected without replacement, and a random variable that counts how many sampled units belong to one chosen category. Getting those ingredients right is more important than memorizing a formula.
“Success” is simply the name for the category being counted. It need not mean a favorable outcome. For instance, a success might be a claim with a coding error, a policy with a particular rider, or a manufactured item that fails inspection. The other category is called failure only for convenience.
A standard hypergeometric model assumes that the population membership and category labels are fixed during sampling and that the sample is selected at random. In a simple random sample, every subset of \(n\) distinct population units has the same chance of being selected. The sample size is set in advance; the process does not stop when a target number of successes is reached.
Setting Up the Count
A reliable setup begins by naming the actual population units. Then define which category will count as success, determine how many units belong to each category, and state the sample size and selection rule. Finally, define the random variable in words. This order prevents common errors such as counting the wrong unit or treating a sample of fixed size as a sequence of independent trials.
Specify what one unit is and how many distinct units are eligible for selection.
State the success definition and count how many population units meet it. The remaining units form the other category.
Give the fixed sample size and verify that units are selected randomly without replacement.
Let the variable equal the number of sampled units in the chosen category. Keep the variable distinct from the labels or order of the selected units.
In concise notation, \(N\) can denote the population size, \(K\) the number of successes in that population, and \(n\) the fixed sample size. These counts help record the setup, but their interpretation must always be stated in context. The random variable is a count, not an indicator for one draw and not the entire ordered sequence of draws.
Worked Examples: Translating Sampling Situations
Worked Example: Auditing a Fixed Set of Claim Files
A hypothetical insurer has 24 distinct claim files in a review batch. Seven files contain a specified coding error, and the other files do not. An auditor selects five distinct files by simple random sampling, without replacement. Define a suitable count and translate the event that at least two selected files contain the error.
Method: identify the population, category, sample design, and count. The population consists of the 24 claim files, so \(N=24\). Define success as “the file contains the specified coding error.” There are \(K=7\) successes. The failure count is \(24-7=17\). The auditor takes a fixed-size sample of \(n=5\) files and does not return a selected file to the batch.
Let \(X\) be the number of selected files that contain the coding error. This is a count among the five sampled files, so the appropriate model is hypergeometric. The population fraction in the success category is \(7/24=0.2917\) to four decimal places; that fraction describes the batch, not an independent success probability that remains unchanged after each selection.
The event “at least two selected files contain the error” is \(X\geq 2\). It asks about the count in the sample, not whether the first two files selected contain errors. No probability calculation is needed to define the event or the model.
Check. The two population categories total \(7+17=24\), and the sample contains five distinct files. The setup avoids the mistake of treating the five selections as independent trials with a fixed success chance.
Worked Example: Counting Policies With a Rider
A hypothetical portfolio contains 18 distinct policies: six include a specified optional rider and 12 do not. An analyst randomly selects four different policies from the portfolio for a file review. Let \(Y\) count policies with the rider. Express each event in terms of \(Y\): (a) none of the reviewed policies has the rider, and (b) at least one has the rider.
Method: fix the counted category, then translate each verbal condition. The population is the 18 policies, not the policyholders or the four review slots. Define success as “policy includes the rider.” There are six successes, since the problem states that six of the policies include it. The other-category count is \(18-6=12\).
The sample size is four, chosen randomly and without replacement. Define \(Y\) as the number of those four policies that include the rider. Since the selection is fixed in size and consists of distinct policies, \(Y\) has a hypergeometric model.
(a) “None” means the count of rider policies is zero, so the event is \(Y=0\). (b) “At least one” means the count is one or more, so the event is \(Y\geq 1\). These are statements about the sample count; they do not say that all four policies share the same status.
Check. The categories total \(6+12=18\), and the four reviewed policies are distinct. The setup avoids counting the number of policyholders in place of the number of policies, which would be wrong if a policyholder could own multiple policies.
Worked Example: Interpreting a Particular Sample
A hypothetical batch contains 20 distinct claim records. Nine records show a positive payment, and the remaining records show a zero payment. Three records are selected randomly without replacement. Let \(Z\) be the number of selected records showing a positive payment. A particular selected set contains one zero-payment record and two positive-payment records. Identify the model and the value of \(Z\) for this selected set.
Method: define the population category before reading the sample outcome. The units are the 20 distinct claim records. Define success as “the record shows a positive payment.” The success count in the population is nine. The failure count is \(20-9=11\), corresponding to the zero-payment records. The fixed sample size is \(n=3\), with no replacement.
The random variable \(Z\) counts the positive-payment records among the three selected records, so \(Z\) follows a hypergeometric model. In the particular set described, two of the three records show positive payments. Therefore, for that set, \(Z=2\).
The count does not depend on the order in which the three records were selected. If the same set were listed as positive, zero, positive or as zero, positive, positive, the count would still be \(Z=2\).
Check. The population categories total \(9+11=20\), and the sample count \(2\) matches the two positive-payment records stated. This avoids confusing the recorded value on each claim with the number of sampled claims in the chosen category.
Why the Draws Are Dependent
In sampling without replacement, observing one unit changes which units remain available. If \(j\) units have already been drawn, and \(s\) of those are successes, then among the \(N-j\) remaining units there are \(K-s\) successes. Conditional on that observed history, the chance that the next draw is a success is therefore
This expression follows by counting the remaining units in the two categories: \(N-j\) total units remain, of which \(K-s\) are successes. Under random selection from those remaining units, the fraction of remaining units that are successes is the next-draw success chance. Because \(j\) and \(s\) can change with what has already been drawn, this chance generally changes from draw to draw.
For example, suppose a finite group has 12 units, of which five are successes. If the first selected unit is a success, then 11 units remain, including four successes. The conditional success chance on the next selection is \(4/11=0.3636\) to four decimal places. Using the original fraction \(5/12=0.4167\) instead would ignore the known first selection. The key distinction is dependence between draws, not merely that the population happens to be finite.
The hypergeometric model is not appropriate if the selection is with replacement and each draw has the same success chance; that design instead gives independent trials with a fixed success probability. Nor is the standard model automatically appropriate if selection is nonrandom, if some units have unequal selection chances, or if the category labels can change during sampling. In each case, the sampling mechanism must be examined before choosing a model.
Common Exam Traps
- Using the wrong unit. If the population is claim files, the count is files with the feature, not the number of people associated with those files.
- Leaving success undefined. “Count the good ones” is ambiguous. State the exact category that makes a unit a success.
- Ignoring the sampling rule. A fixed sample size alone is not enough. Verify both random selection and no replacement.
- Treating a count as a sequence. The random variable records how many sampled units are successes; it does not record their positions or order.
- Assuming an unchanged success chance. With no replacement, the remaining category counts depend on earlier selections. A fixed fraction \(K/N\) does not describe every conditional draw.
Check Your Understanding
For each situation, identify the population categories, state the sample size and selection rule, and define or interpret the count as requested.
- A hypothetical batch has 16 devices, five of which fail a specified test. Three distinct devices are randomly selected without replacement. Define \(X\) as the number that fail. (Answer: \(X\) counts failures among the three; the population has 5 failures and 11 nonfailures, with fixed sample size 3 and no replacement.)
- A hypothetical book of 22 policies contains eight policies with monthly billing and 14 without it. A random sample of six distinct policies is reviewed. Let \(Y\) count monthly-billed policies. Express “exactly two reviewed policies use monthly billing” using \(Y\). (Answer: \(Y=2\); \(Y\) is a hypergeometric count with population categories 8 and 14 and sample size 6.)
- A sealed tray has 30 distinct applications, 12 of which are incomplete. Four are sampled randomly without replacement. Let \(W\) be the number incomplete. Express “at most one is incomplete” using \(W\). (Answer: \(W\leq 1\); the categories are 12 incomplete and \(30-12=18\) complete, with sample size 4.)
- A group contains 15 distinct records, four marked for a secondary review. Two records are selected at random without replacement, and the first selected record is known to be marked. How many marked records remain among the 14 available, and what is the conditional chance that the second selection is marked? (Answer: 3 marked records remain; the conditional chance is \(3/14\).)
- A hypothetical inventory has 40 items, including 9 with a specified defect. A reviewer selects five distinct items by judgment rather than by a random procedure. Is the standard hypergeometric model justified as stated? Give the key reason. (Answer: No. The population counts and no-replacement feature are specified, but judgment selection does not establish random sampling with equally likely sample subsets.)