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Exam P · Integrated Probability Problem Solving · Tutorial 882 of 915

Distinguishing Stopping-Time And Sampling Models

Choose the correct model from the sampling plan and stopping rule, then calculate probabilities without confusing fixed sample sizes with outcome-driven stopping.

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What You'll Learn

  • Identify whether a process has a fixed number of trials or stops when a target outcome occurs.
  • State the assumptions required for geometric and negative binomial models.
  • Distinguish independent Bernoulli trials from sampling without replacement.
  • Calculate probabilities using binomial and hypergeometric models.
  • Recognize when a sequential sample without replacement is not geometric.
  • Avoid common errors involving the final success and changing conditional probabilities.

Start With the Sampling Plan

Several probability models count successes, but they answer different questions. A binomial model describes the number of successes in a fixed number of independent trials. A geometric or negative binomial model describes how long it takes to reach a specified success. A hypergeometric model counts successes in a fixed sample drawn without replacement from a finite population. The choice depends on the process that generates the data, not just on the fact that successes and failures are being counted.

A reliable first step is to ask two questions: Is the number of trials fixed in advance, or does it depend on observed outcomes? And does each trial have the same success probability independently, or does sampling without replacement change the probabilities? These questions separate the models and reveal when none of the four applies exactly.

Definition: A fixed-sample model specifies the number of trials before observing their outcomes. A stopping-time model continues trials until a rule based on observed outcomes is met; the total number of trials is then random. Sampling with replacement restores the sampled item before the next draw, while sampling without replacement removes it, changing the composition of the remaining population.

Four Models and Their Conditions

The binomial model requires a fixed number \(n\) of trials, two outcomes per trial, a constant success probability \(p\), and independence between trials. If \(X\) is the number of successes, then \(X\sim\operatorname{Bin}(n,p)\), with

$$ P(X=k)=\binom{n}{k}p^k(1-p)^{n-k},\qquad k=0,1,\ldots,n. $$

A binomial model is exact for independent trials with the same success probability, such as draws with replacement from a well-defined population. It is generally not exact for sampling without replacement from a finite population, because earlier draws affect the probabilities of later draws.

The geometric model assumes independent trials, each with the same success probability \(p\), and stops at the first success. Here \(X\) denotes the number of trials through and including the first success, so \(X=1,2,\ldots\). Its probability function is

$$ P(X=k)=(1-p)^{k-1}p,\qquad k=1,2,\ldots. $$

Some settings instead count the number of failures before the first success. That is a different random variable, with support \(0,1,2,\ldots\). Always identify what the question counts before using a geometric formula.

The negative binomial model generalizes the geometric model. It assumes independent trials with the same success probability \(p\), and stops at the \(r\)th success, where \(r\) is fixed. If \(T\) is the number of trials through and including that success, then

$$ P(T=t)=\binom{t-1}{r-1}p^r(1-p)^{t-r},\qquad t=r,r+1,\ldots. $$

The hypergeometric model applies when \(n\) objects are sampled without replacement from a finite population of size \(N\), which contains \(K\) successes and \(N-K\) failures. If \(X\) is the number of successes in the sample, then

$$ P(X=k)=\frac{\binom{K}{k}\binom{N-K}{n-k}}{\binom{N}{n}}, $$

for feasible values of \(k\): \(0\leq k\leq K\), \(0\leq n-k\leq N-K\), and \(0\leq k\leq n\). The sample size \(n\) is fixed; the formula counts the ways to choose \(k\) successes and \(n-k\) failures from the population, divided by the number of possible samples.

Why the Stopping-Time Formulas Work

For the geometric model, the event \(X=k\) requires \(k-1\) failures followed by a success. Independence and the constant success probability give probability \((1-p)^{k-1}p\). For the negative binomial model, \(T=t\) requires the \(t\)th trial to be a success, and exactly \(r-1\) successes among the first \(t-1\) trials. There are \(\binom{t-1}{r-1}\) ways to place those earlier successes. Each qualifying sequence has probability \(p^r(1-p)^{t-r}\), giving the formula.

$$ P(T=t)= \binom{t-1}{r-1} \underbrace{p^{r-1}(1-p)^{t-r}}_{\text{first }t-1\text{ trials}} \underbrace{p}_{\text{success on trial }t} = \binom{t-1}{r-1}p^r(1-p)^{t-r}. $$

The final trial must be a success. The combination therefore chooses the other \(r-1\) success locations from the first \(t-1\) trials, not all \(r\) success locations from \(t\) trials. This detail is a frequent source of incorrect answers.

Worked Example: Waiting for the First Successful Test

A hypothetical calibration procedure tests units one at a time. Each test independently passes with probability \(0.25\), and testing stops at the first pass. Let \(X\) be the number of tests, including the passing test. Find the probability that the first pass is on test 3 and the probability that it occurs by test 3.

Method and assumptions. The number of tests is determined by the first pass, so this is a geometric stopping-time model with \(p=0.25\) and failure probability \(1-p=0.75\). The stipulated independence and constant pass probability are essential; there is no sampling without replacement.

For the first pass on test 3, the first two tests must fail and the third must pass:

$$ P(X=3)=(0.75)^2(0.25)=0.5625(0.25)=0.140625. $$

The first pass occurs by test 3 if it occurs on test 1, 2, or 3. Equivalently, it does not occur by test 3 only if all three tests fail:

$$ P(X\leq 3)=1-(0.75)^3=1-0.421875=0.578125. $$

Rounded to four decimal places, the answers are \(P(X=3)=0.1406\) and \(P(X\leq3)=0.5781\). As a check, the probability of a first pass on test 1 or 2 is \(0.25+0.75(0.25)=0.4375\); adding \(0.140625\) gives \(0.578125\). The method avoids treating the stopping count as a fixed sample size.

Worked Example: Stopping at the Third Success

A hypothetical quality audit records each inspected component as acceptable with probability \(0.30\), independently of all other components. Inspection stops when the third acceptable component is found. Let \(T\) be the total number inspected. Find \(P(T=7)\) and \(P(T\leq7)\).

Method and assumptions. This is a negative binomial model with \(r=3\), \(p=0.30\), and \(1-p=0.70\). The inspection count is not fixed: it ends at the third success.

For \(T=7\), the seventh component must be acceptable, and exactly two of the first six must be acceptable. Thus

$$ \begin{aligned} P(T=7) &=\binom{6}{2}(0.30)^3(0.70)^4\\ &=15(0.027)(0.2401)\\ &=15(0.0064827)=0.0972405. \end{aligned} $$

The event \(T\leq7\) means that at least three acceptable components appear in the first seven. The complement is zero, one, or two acceptable components:

$$ \begin{aligned} P(T\leq7) &=1-\left[(0.70)^7+\binom{7}{1}(0.30)(0.70)^6 +\binom{7}{2}(0.30)^2(0.70)^5\right]\\ &=1-(0.0823543+0.2470629+0.3176523)\\ &=1-0.6470695=0.3529305. \end{aligned} $$

Rounded to four decimal places, \(P(T=7)=0.0972\) and \(P(T\leq7)=0.3529\). As a check, \(T=7\) is one way for the third success to have arrived by trial 7, so its probability \(0.0972405\) is smaller than \(P(T\leq7)=0.3529305\). A common error is to use \(\binom{7}{3}\) for \(P(T=7)\); that incorrectly allows the seventh trial to be a failure, although stopping at the third success on trial 7 requires a success on trial 7.

Worked Example: A Fixed Number of Independent Trials

A hypothetical laboratory processes eight samples. Each sample independently triggers a particular alert with probability \(0.25\). Let \(X\) be the number of alerts. Find the probability of at least two alerts.

Method and assumptions. The laboratory processes exactly \(n=8\) samples, regardless of the alerts, and the alert outcomes are independent with common probability \(p=0.25\). Thus \(X\sim\operatorname{Bin}(8,0.25)\); this is not a stopping-time model.

It is simpler to subtract the probabilities of zero and one alert from 1:

$$ \begin{aligned} P(X\geq2) &=1-P(X=0)-P(X=1)\\ &=1-(0.75)^8-\binom{8}{1}(0.25)(0.75)^7\\ &=1-0.1001129150-0.2669677734\\ &=0.6329193116. \end{aligned} $$

Rounded to four decimal places, \(P(X\geq2)=0.6329\). As a check, the three disjoint cases \(X=0\), \(X=1\), and \(X\geq2\) have probabilities summing to \(0.1001129150+0.2669677734+0.6329193116=1\). The fixed sample size, rather than the alert count, determines that the binomial model is appropriate.

Worked Example: Sampling Without Replacement

A hypothetical shipment contains 12 circuit modules, of which 4 are faulty and 8 are functioning. An auditor selects 3 modules without replacement. Find the probability of exactly 2 faulty modules and the probability of at least 1 faulty module.

Method and assumptions. The sample size is fixed at \(n=3\), but selection is without replacement from a finite population with \(N=12\) and \(K=4\) faulty modules. Use the hypergeometric model.

For exactly 2 faulty modules, choose 2 of the 4 faulty modules and 1 of the 8 functioning modules, then divide by the number of ways to choose any 3 of the 12:

$$ P(X=2)=\frac{\binom{4}{2}\binom{8}{1}}{\binom{12}{3}} =\frac{6(8)}{220} =\frac{12}{55} \approx 0.2182. $$

For at least 1 faulty module, use the complement that all 3 selected modules function:

$$ P(X\geq1)=1-\frac{\binom{8}{3}}{\binom{12}{3}} =1-\frac{56}{220} =\frac{41}{55} \approx 0.7455. $$

The answers, rounded to four decimal places, are \(0.2182\) and \(0.7455\). As a check, the complement probability is \(56/220\approx0.2545\), and \(0.7455+0.2545=1.0000\). The method avoids assuming that each draw still has faulty probability \(4/12\).

When Stopping and Sampling Without Replacement Interact

A stopping rule does not by itself imply a geometric or negative binomial model. Those models require independent trials with a constant success probability. For example, suppose objects are drawn without replacement and drawing stops at the first marked object. The probability of a mark on the next draw changes as unmarked objects are removed. The stopping count is random, but the trials are dependent, so the geometric formula does not apply.

Worked Example: A Stopping Rule Without Replacement

A hypothetical container holds 12 components: 4 marked and 8 unmarked. Components are drawn without replacement until the first marked component appears. Find the probability that the first marked component is drawn third.

Method and assumptions. This is a sequential calculation, not a geometric model: the marked proportion changes after each unmarked draw. For the first marked component to appear third, the first two draws must be unmarked and the third marked:

$$ P(\text{first mark on draw 3}) =\frac{8}{12}\cdot\frac{7}{11}\cdot\frac{4}{10} =\frac{224}{1320} =\frac{28}{165} \approx0.1697. $$

The factors use the updated population after each draw. As a check, the first two unmarked draws have probability \((8/12)(7/11)=14/33\), and then the probability of a mark is \(4/10\), giving \((14/33)(4/10)=28/165\). A geometric calculation \( (2/3)^2(1/3)=4/27\approx0.1481\) is wrong because it keeps the initial marked proportion \(1/3\) unchanged.

A Practical Model-Selection Check

Before calculating, describe both the count and the experiment in words. A fixed number of independent trials points to a binomial count; a fixed number of draws without replacement from a finite population points to a hypergeometric count. A process that stops at the first success points to a geometric model only when trials are independent and have constant success probability. Stopping at the \(r\)th success points to a negative binomial model under the same assumptions.

The distinction between stopping and fixed sampling is particularly important when a question gives a phrase such as “until,” “exactly \(n\) are selected,” or “a sample of \(n\) is inspected.” “Until” commonly signals a random trial count, while “a sample of \(n\)” commonly signals a fixed count. Neither phrase determines independence by itself: check whether the process replaces each selected item or otherwise specifies independent trials.

  • Fixed count, independent, common \(p\): binomial number of successes.
  • Stop at first success, independent, common \(p\): geometric number of trials.
  • Stop at the \(r\)th success, independent, common \(p\): negative binomial number of trials.
  • Fixed sample, finite population, no replacement: hypergeometric number of successes.
  • Stopping with changing probabilities: do not force a geometric or negative binomial formula; calculate from the sequential probabilities or use a suitable model.

A frequent exam trap is to identify a distribution solely from the words “successes in trials.” First determine what is fixed and what is random. Then check independence and whether the success probability remains constant. This disciplined model check is aligned with the SOA Exam P (Probability) syllabus.

Check Your Understanding

For each hypothetical setting, identify the appropriate model and calculate the requested probability. Give probabilities to four decimal places.

  1. A device is tested until its first successful startup. Startup trials are independent, with success probability \(0.40\). Find the probability the first success is on test 4. (Answer: geometric; \(0.0864\))
  2. Six independent sensor readings each exceed a threshold with probability \(0.30\). Find the probability that exactly two readings exceed it. (Answer: binomial; \(0.3241\))
  3. A process continues until its third successful calibration; each calibration independently succeeds with probability \(0.20\). Find the probability the third success occurs on trial 6. (Answer: negative binomial; \(0.0410\))
  4. A crate has 15 valves, 5 of which are defective. Four are chosen without replacement. Find the probability exactly two are defective and the probability at least one is defective. (Answer: hypergeometric; \(0.3297\) and \(0.8462\))
  5. A box contains 10 parts, 3 marked. Parts are drawn without replacement until the first marked part is found. Find the probability the first marked part is drawn third. (Answer: sequential without replacement, not geometric; \(0.1750\))

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