The Gold Standard of Probability | Sampling Without Replacement and the First Defect

Begin with twelve distinct labelled components: three defective and nine sound. The labels let us distinguish individual objects even when they share the same status. Each inspection selects one of the remaining components uniformly and removes it from further selection.

There are several legitimate questions. What is the chance the first inspected component is defective? How many defects will a four-item sample contain? On which inspection will the first defect appear? Will a capped inspection find any defect at all?

These questions use the same box but different random variables. Let X count defects in a fixed four-item sample. Let T count inspections through the first defect when the process continues until one appears. X is a count in a preselected sample size; T is a waiting position.

Write the sampling rule beside either variable. With replacement would mean putting each component back before the next selection. That creates a different experiment. Assuming a well-randomised reset, the original one-in-four probability could then remain constant across independent selections.

Without replacement, we need the current remaining counts after the observed history. The arithmetic is not harder because the objects are called components. It is different because the experiment changes the population available for the next draw.

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Before inspection, P(first defective) = 3/12 = 0.25. If the first component is sound, eleven remain and all three defects remain. The next conditional probability is 3/11, approximately 27.27%.

If the first component is defective instead, eleven remain with two defects. Now the next conditional probability is 2/11, approximately 18.18%. Knowing what happened changes the remaining composition.

The last row is especially useful. After three sound inspections, the next defect probability is one third. The object has not developed a defect because we looked at other objects. The information and remaining collection have changed.

A common mistake is reducing the denominator while leaving the numerator unchanged after every possible result. That works only along a sequence of sound removals. After a defect is removed, both the total count and defect count decrease.

Use a small state record: sound remaining, defective remaining, total remaining. Update the state first, then write the next fraction. This makes the reason for every branch visible.

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There is a subtle point behind the changing fractions. Before observing any outcomes, the second position in a uniformly random ordering still has defect probability 3/12. Every position has the same marginal defect probability by symmetry.

That does not make the draws independent. Once we learn the first result, the second probability changes to 3/11 or 2/11. A marginal probability and a probability conditional on observed history answer different questions.

Check with the two first-result paths. The chance of first defective and second defective is (3/12)(2/11). The chance of first sound and second defective is (9/12)(3/11). Adding them gives (6+27)/132 = 33/132 = 1/4.

This calculation explains how a constant marginal probability can coexist with dependence. It would be inaccurate to say that every unobserved position becomes intrinsically more likely to be defective merely because its position number is larger.

The conditional probability guide develops the meaning of the information after the vertical bar. Here the practical phrase is “given the observed removals”.

If a learner says the probability is both one quarter and one third, ask which information each statement assumes. Sometimes the apparent contradiction disappears when the question is stated completely.

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For the first defect to occur on inspection four, the first three selections must be sound and the fourth defective. Follow that exact path:

P(T = 4) = (9/12)(8/11)(7/10)(3/9) = 7/55 ≈ 0.127273.

Each fraction comes from a different remaining box. The three defects remain available during the three sound removals, while the number of sound components falls from nine to eight to seven. At the fourth selection, nine objects remain with three defects.

Do not multiply by four to allow the defect to occur in another position. The question fixes its first position at four. A fixed-size sample containing exactly one defect permits several arrangements and is a different event.

An alternative counting argument starts with a random ordering of all twelve labels. The three defect positions form an equally likely three-position subset, giving C(12,3) = 220 possibilities. If the first defect is fourth, the other two defect positions must lie among positions five through twelve: C(8,2) = 28 possibilities. Thus 28/220 = 7/55.

Here C(n,k) means the number of k-element selections from n distinct positions. The sequence method and position-counting method agree because both represent the same experiment.

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For a four-item sample to contain no defects, all four selected components must be sound. The sequential probability is (9/12)(8/11)(7/10)(6/9) = 14/55 ≈ 0.254545.

Therefore the probability of at least one defect in four selections is 1−14/55 = 41/55 ≈ 0.745455, or 74.55%.

The same complement can be counted without ordering. There are C(12,4) = 495 possible four-component subsets and C(9,4) = 126 all-sound subsets. Their ratio is 126/495 = 14/55.

Now compare the independent reset model. With a constant 0.25 defect probability per independent attempt, no defect in four would have probability 0.75^4 = 0.31640625. Detection by four would be about 68.36%, rather than 74.55%.

Both calculations are valid for their own experiments. The difference is not a rounding error. Without replacement, each sound removal leaves the known defects concentrated within a smaller remaining group.

The earlier first-success and retry-limit guide uses independent reset trials. Reading the two cases together shows exactly which condition changes and why the old multiplication cannot be copied unchanged.

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Let X count defects in a random sample of four components. X can be zero, one, two or three; four is impossible because the box contains only three defects.

To count samples with x defects, choose x of the three defects and 4−x of the nine sound components. Divide by all 495 equally likely four-item subsets:

P(X = x) = C(3,x)C(9,4−x)/C(12,4).

The counts total 495. Their probabilities therefore total one. The weighted average defect count is (252 + 216 + 27)/495 = 1.

This is a hypergeometric model: a fixed-size random sample drawn without replacement from a finite population with a specified composition. OpenStax’s hypergeometric chapter provides the formal distribution background.

Check possible values before using a calculator. A numerical routine that is fed the wrong population count or an impossible sample event cannot determine what the problem intended. The physical box supplies those constraints.

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In the independent four-trial model with p = 0.25, the expected defect count is also one. That agreement does not make its distribution identical to the finite-box distribution.

For the independent binomial count, variance is 4(0.25)(0.75) = 0.75. For the stated simple random sample without replacement, the finite-population factor is (12−4)/(12−1) = 8/11. Multiplying gives variance 0.75(8/11) = 6/11 ≈ 0.545455.

We can verify the finite-box variance directly from the table rather than relying on the formula. E(X²) = [1(252)+4(108)+9(9)]/495 = 765/495 = 17/11. Subtracting E(X)² = 1 gives 6/11.

The smaller variance reflects the finite composition constraint in this model. Finding a defect leaves fewer defects for later selections. It is not a universal claim that every dependent process has less variability than an independent one.

For learning, the important result is that equal means do not establish equal probabilities of zero, one or several defects. The question may depend on a tail or an event probability that the mean alone does not preserve.

Keep the model, sample fraction and distribution beside the summary. A headline “one defect expected” cannot answer whether at least one defect will be found in this particular sample.

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Return to T, the inspection position of the first defect. Its possible values run from one through ten. The first nine components could all be sound, but after those nine removals a defect must appear next under the known-composition, perfect-inspection model.

An unlimited geometric waiting time has no such finite upper bound. Substituting p = 0.25 into E(T)=1/p would give four, but that geometric formula belongs to independent constant-probability trials.

A small enumeration makes the difference visible without advanced notation. Take five objects with two defects. The ten equally likely pairs of defect positions have earliest positions distributed as follows: four pairs begin at one, three begin at two, two begin at three and one begins at four.

The mean first-defect position is [1(4)+2(3)+3(2)+4(1)]/10 = 2. The independent geometric mean with p = 2/5 would be 2.5. Different remaining-population rules create different waiting behaviour.

In the twelve-object example, a four-inspection cap may stop without finding a defect. Record “four inspected, none detected”, rather than inventing a first-defect time of four. The cap has not changed the meaning of T; it has limited what we observe about it.

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Our teaching box contains three defects by assumption. A four-item all-sound sample still occurs with probability 14/55. The observation is compatible with remaining defects because many components were not inspected.

In an actual investigation, the total number of defects is often unknown. One can compare different candidate population compositions or use an appropriate inferential method. It would be circular to treat the assumed count used in a teaching calculation as a measured fact about a real batch.

Inspection also may not classify every item perfectly. If defects can be missed or sound components can be flagged, finding a defect in the selected sample and observing a defect report are different events. Those classification rates require their own evidence.

Similarly, uniform random selection matters. Picking the easiest items to reach may not support the same probability model as selecting a random subset of labels. Physical access, clustering and selection practices can change what the sample represents.

For these reasons, this article does not give an acceptance threshold or a recommendation about a real product. The useful capability is recognising which assumptions made the number possible and which additional evidence an actual decision needs.

A careful report can be modest and informative: “Four randomly selected items were inspected under this method and no defect was reported.” That preserves the observation without replacing the uninspected population with an unsupported guarantee.

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Use twelve labelled cards, with three marked on their reverse sides. A paper exercise can supply their random ordering; no physical inspection equipment is needed. Reveal cards one by one without returning them.

The marked cards replace defective components. The first marked position and the count in four reveals have the same mathematical structure, provided the ordering is uniformly random and the marking is identified correctly.

Now change one rule. Return each revealed card and randomise again. The experiment becomes a reset model, and the earlier finite-box probabilities no longer apply. Change another rule: preferentially choose larger cards. The equally likely subset count may now be unsuitable.

Change the recording question as well. “Stop at the first marked card” does not produce the same observed dataset as “reveal exactly four cards”. If only first-mark positions are retained, the record does not directly list the total number of marked cards that would occur in every four-card sample.

Ask the learner what stayed the same: labels, population composition and the event of interest. Then ask what changed: replacement, selection probabilities or stopping. This is a more useful transfer check than simply renaming components as cards while keeping every instruction identical.

The mathematics travels through the experiment’s structure. A familiar story alone does not establish that the same model belongs there.

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Update after a sound result A box contains eight items with two defects. After one sound removal, the next defect probability is 2/7. After one defective removal, it is 1/7. Update the remaining composition before choosing the fraction.

Find any defect in two selections For that same box, no defect has probability (6/8)(5/7) = 15/28. At least one defect therefore has probability 13/28. The complement includes both one-defect and two-defect outcomes without double counting.

First defect exactly second The probability is (6/8)(2/7) = 3/14. Do not add the path beginning with a defect; that would contradict “first defect second”.

Inspect every item If at least one defect is known to be present and inspection is perfect, selecting all items must find a defect. A model result below one for complete inspection signals that the replacement or population assumptions were mishandled.

Name the denominator “After three sound inspections in the twelve-item example, one third are defective” refers to the nine remaining items, not the original twelve or the already inspected three. The words identify which collection the fraction describes.

Identify the missing record A capped run with no defect gives T greater than the cap. It does not reveal T’s eventual exact value. Preserve the number inspected, the outcomes, whether the cap was reached and the rule used.

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Does sampling without replacement always make the next probability increase? No. A sound removal increases the next defect probability in this known-composition model; a defect removal decreases it. Before any results are known, each position still has the same marginal probability by symmetry.

Is a hypergeometric model about the first success? The usual hypergeometric variable counts group members in a fixed-size sample. A first-success position is a related waiting-time question. Define the variable instead of choosing a distribution from the story alone.

When is a binomial approximation acceptable? It may be useful when the sampling fraction is small and the intended approximation is accurate enough for the task. Check the relevant event and error tolerance; no universal sample-fraction shortcut guarantees every tail probability is adequate.

Sources and scope The linked OpenStax hypergeometric chapter supports the finite-population model. All box sizes, counts, sample tables, waiting-position enumeration and practice calculations here are original teaching examples derived from the stated assumptions.

Follow the probability structure Return to The Gold Standard of Probability, continue with conditional probability, and compare the independent experiment in First Success, Retry Limits and Unfinished Runs.

Connect the learning Learning That Transfers into Life explains why a changed selection rule is a useful test of understanding. The How X Works hub connects the wider library.

Before using the next probability fraction, describe what remains available for selection. That short account often reveals which numerator, denominator and model the question actually needs.

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