Simple Random Sampling Without Replacement

Why the Sampling Rule Matters

Suppose a clinic wants feedback from a fixed list of patients and plans to survey only some of them. Before interpreting their answers, it matters how the patients were chosen. A probability sample uses a random selection rule with known chances of selection. A simple random sample without replacement is a particularly clear design: it gives each possible sample of the planned size the same chance of being selected, and no unit can be selected more than once.

The “without replacement” part makes selection chances change as units are chosen. It is therefore important to distinguish the chance that one named unit is included from the chance that a particular group of units appears together. Both probabilities follow from the same precisely stated sampling rule.

Definition: Consider a finite population of \(N\) distinct units, and choose a sample of fixed size \(n\), where \(1 \leq n \leq N\). A simple random sample without replacement (SRSWOR) is a random sample in which every subset of \(n\) different population units has the same probability of selection. Each unit can appear at most once in the sample.

The population is treated as fixed: its \(N\) units are the units eligible for selection, and the randomness lies in which subset is chosen. The design assumes that the units can be distinguished and that the selection mechanism really gives every size-\(n\) subset the same chance. In an actual survey, the sampling frame must correspond adequately to the target population; the probability calculations here describe selection from the stated finite list.

Equal Chances: Units and Whole Samples

There are \(\binom{N}{n}\) different subsets of size \(n\) that could be selected from \(N\) units. Under SRSWOR, each such subset has probability \(1/\binom{N}{n}\). This is a statement about whole samples, not just about individual units. By symmetry, each individual unit also has the same probability of inclusion, but that probability is generally not \(1/\binom{N}{n}\).

Theorem: In an SRSWOR of size \(n\) from a population of \(N\) units, the probability that any specified unit is included is \(n/N\). The probability that any particular subset of \(n\) units is selected is \(1/\binom{N}{n}\).

These are two different probabilities. A unit can appear in many possible samples, so its inclusion probability is the number of samples containing it divided by the total number of possible samples. A particular sample, by contrast, is only one of those possibilities.

Worked Example: The Chance That One Patient Is Included

A clinic has a fixed list of 8 eligible patients and chooses an SRSWOR of 3 patients. What is the probability that a specified patient, such as patient A, is included?

There are \(\binom{8}{3}=56\) possible samples. If patient A is included, the other 2 sample places must be filled by 2 of the remaining 7 patients, giving \(\binom{7}{2}=21\) samples that contain A. Therefore,

P(\text{A is included}) =\frac{\binom{7}{2}}{\binom{8}{3}} =\frac{21}{56} =\frac{3}{8} =0.375.

The same result follows from \(n/N=3/8\). Thus each named patient has a 0.375 probability of being included. This does not mean that the probability of any particular three-patient sample is 0.375: each particular sample has probability \(1/56\).

The calculation can be generalized. There are \(\binom{N}{n}\) equally likely samples in total. For a specified unit to be included, choose the other \(n-1\) units from the remaining \(N-1\); there are \(\binom{N-1}{n-1}\) such samples. Hence

P(\text{specified unit included}) =\frac{\binom{N-1}{n-1}}{\binom{N}{n}} =\frac{n}{N}.

The last equality follows by writing the binomial coefficients using factorials and cancelling: \(\binom{N-1}{n-1}/\binom{N}{n} =[(N-1)!/((n-1)!(N-n)!)]/[N!/(n!(N-n)!)] =n/N\).

Worked Example: The Chance of One Particular Sample

A department has 12 instructors and selects 4 for a short planning meeting using SRSWOR. What is the probability that the selected sample is exactly instructors A, B, C, and D? Also, what is the probability that instructor A is included?

The number of possible samples is \(\binom{12}{4}=12!/(4!8!)=495\). Since the specified set \(\{A,B,C,D\}\) is one of those 495 equally likely samples, its probability is

P(\{A,B,C,D\}\text{ is selected}) =\frac{1}{\binom{12}{4}} =\frac{1}{495} \approx 0.002020.

For instructor A to be included, the other 3 places can be filled by any 3 of the 11 other instructors. There are \(\binom{11}{3}=165\) such samples, so

P(\text{A is included}) =\frac{165}{495} =\frac{4}{12} =\frac{1}{3} \approx 0.3333.

The probability of a particular sample is much smaller than the probability of including A because many different samples contain A. Both calculations rely on all 495 samples being equally likely.

Selection Without Replacement and Dependence

When units are selected without replacement, a unit already selected cannot be selected again. Consequently, the inclusion status of one unit affects the chance that another unit is included. For example, if a sample has room for 3 of 8 units and one specified unit is known to be included, only 2 places remain among the other 7 units.

Theorem: For two distinct specified units in an SRSWOR from a population of size \(N\), with \(N \geq 2\) and \(1 \leq n \leq N\), the probability that both are included is \(n(n-1)/[N(N-1)]\). The conditional probability that the second specified unit is included, given that the first is included, is \((n-1)/(N-1)\).

To see the conditional result, once the first specified unit is known to be included, the remaining \(n-1\) sample places are occupied by units from the other \(N-1\) population units. By symmetry, each of those \(N-1\) units has the same conditional chance of occupying one of those places. The chance for the second specified unit is therefore \((n-1)/(N-1)\). This reasoning also covers \(n=1\), when there are zero remaining places and the conditional chance is 0. The condition \(N \geq 2\) is necessary because two distinct specified units cannot be considered when the population has only one unit.

The joint probability follows by multiplying the probability that the first unit is included by the conditional probability that the second is included:

P(\text{both included}) =\frac{n}{N}\cdot\frac{n-1}{N-1} =\frac{n(n-1)}{N(N-1)}.

For a sample of size \(n=1\), this expression gives 0, as two distinct units cannot both be in a one-unit sample. When \(n=N\), it gives 1, because a census includes every unit.

Worked Example: Two Specified Units and a Conditional Chance

A fixed population has 8 patients, and an SRSWOR of 3 is selected. Find (a) the probability that patients A and B are both included, and (b) the probability that B is included, given that A is included.

Here \(N=8\) and \(n=3\), with \(N\geq2\) and \(1\leq n\leq N\). For the joint probability,

P(\text{A and B included}) =\frac{3}{8}\cdot\frac{2}{7} =\frac{6}{56} =\frac{3}{28} \approx 0.1071.

For the conditional probability, after A is known to be included, 2 sample places remain among the other 7 patients:

P(\text{B included}\mid\text{A included}) =\frac{3-1}{8-1} =\frac{2}{7} \approx 0.2857.

The unconditional inclusion probability for B is \(3/8=0.375\), which is larger than \(2/7\). Knowing that A is in the sample leaves fewer places for the other patients, so B’s conditional chance decreases.

More generally, when \(1 \leq n < N\), the conditional chance \((n-1)/(N-1)\) is less than the unconditional chance \(n/N\). Indeed,

\frac{n-1}{N-1}<\frac{n}{N} \quad\Longleftrightarrow\quad N(n-1)<n(N-1) \quad\Longleftrightarrow\quad n<N.

Thus inclusion indicators for distinct units are not independent in a non-census SRSWOR. “Not independent” is a precise statement about probabilities: knowing that one unit is included changes the probability that another is included. At \(n=N\), all units are included with certainty, so the conditional and unconditional inclusion probabilities are both 1.

Probabilities for Groups of Units

The equal-subset rule also makes it possible to calculate chances involving a group, such as whether a sample contains at least one patient from a particular category. A useful strategy is often to count the samples that do not meet the condition, then subtract that probability from 1.

Worked Example: At Least One Unit From a Group

A population contains 10 units, of which 4 belong to a specified group and 6 do not. An SRSWOR of 3 units is selected. Find the probability that the sample includes at least one member of the group.

There are \(\binom{10}{3}=120\) possible samples. A sample with no member of the specified group must choose all 3 units from the 6 outside the group. There are \(\binom{6}{3}=20\) such samples. Therefore,

P(\text{no group members included}) =\frac{\binom{6}{3}}{\binom{10}{3}} =\frac{20}{120} =\frac{1}{6}.

The event “at least one group member is included” is the complement of “no group members are included.” Thus,

P(\text{at least one group member included}) =1-\frac{1}{6} =\frac{5}{6} \approx 0.8333.

This calculation uses the uniform probability of each size-3 subset. It does not treat the three selections as independent draws; the sample is chosen without replacement.

What the Definition Does—and Does Not—Guarantee

SRSWOR is a design statement about how a sample is selected. It guarantees equal probabilities for all size-\(n\) subsets, and therefore equal inclusion probabilities for individual units. It does not guarantee that the sample will resemble the population on every characteristic, especially when the sample is small. Nor does the design by itself guarantee accurate measurement or responses from every selected unit.

It is also important not to reverse the definition. Equal inclusion probabilities for individuals alone do not necessarily mean that all subsets are equally likely. For example, suppose a population contains four units, A, B, C, and D, and a size-2 design selects either \(\{A,B\}\) or \(\{C,D\}\), each with probability \(1/2\). Each unit has inclusion probability \(1/2\), but the other four possible pairs have probability 0. This is not an SRSWOR because its possible samples are not all equally likely.

Key takeaway: An SRSWOR gives every fixed-size subset the same selection probability. Each unit’s inclusion chance is \(n/N\), while the chance of joint inclusion reflects the limited number of sample places and is not obtained by assuming independent selections.

Common Mistakes

Confusing a sample’s probability with a unit’s inclusion probability. A particular subset has probability \(1/\binom{N}{n}\); a specified unit has probability \(n/N\).
Assuming selection events are independent. Without replacement, the chance that a second unit is included, given the first is included, is \((n-1)/(N-1)\), not generally \(n/N\).
Calling any equal-chance design an SRSWOR. Equal inclusion chances for individuals are not enough; all subsets of the required size must be equally likely.
Using the two-unit conditional formula when \(N=1\). The expression \((n-1)/(N-1)\) requires \(N\geq2\), because it refers to two distinct units.
Assuming a random sample must match the population exactly. Random selection gives a defined probability design; it does not ensure that every characteristic is perfectly balanced in the realized sample.

Check Your Understanding

Use the definition of SRSWOR and the probability rules developed above.

A population has 15 units, and an SRSWOR of 5 is selected. What is the inclusion probability for a specified unit?
For the same population and sample size, what is the probability of one particular five-unit subset?
A population has \(N=9\), and an SRSWOR of \(n=4\) is selected. Find the probability that two specified distinct units are both included.
Explain why, for a non-census sample, knowing that one specified unit is included changes the chance that another specified unit is included.
In a design with 6 units, a size-2 sample is selected only from the pairs \(\{A,B\}\), \(\{C,D\}\), and \(\{E,F\}\), each with probability \(1/3\). Does equal inclusion probability for each unit make this an SRSWOR? Explain.