Why Waiting-Time Models Need Assumptions
A waiting-time model describes how many trials are needed before a specified success occurs. For example, an analyst might count reviewed records until the first record meeting a criterion, or inspections until the third acceptable component is found. The fact that trials repeat does not, by itself, justify a geometric or negative binomial model. The probability mechanism must remain suitable from one trial to the next.
The foundation is a sequence of independent, identically distributed Bernoulli trials. “Success” is a label for the outcome of interest; it need not mean a favorable result. A trial might be a flagged claim, a defective component, or a policyholder meeting a specified condition. The event must be defined before assigning its probability.
For a finite-valued waiting time until success, the parameter must satisfy \(0<p\leq 1\). When \(0<p<1\), both outcomes are possible. When \(p=1\), every trial succeeds, so the waiting time is degenerate: the first success occurs on the first trial, and the \(r\)th success occurs on the \(r\)th trial. If \(p=0\), success never occurs. The waiting time is infinite, not a standard finite-valued geometric or negative binomial random variable.
When the assumptions hold, let \(T\) denote the trial number of the first success. Then \(T\) has a geometric distribution with parameter \(p\). If \(r\) is a fixed positive integer and \(T_r\) denotes the trial number of the \(r\)th success, then \(T_r\) has a negative binomial waiting-time model with parameters \(r\) and \(p\). The trial number includes the success-producing trial.
Checking the Bernoulli Trial Conditions
Before selecting a waiting-time model, identify the trial, the success event, and the probability mechanism. A useful check is to ask three separate questions:
- Is each trial Bernoulli? The defined event must have two outcomes for the analysis: success or failure. Several recorded outcomes can be grouped into success and its complement if the question defines them that way.
- Is the success probability fixed? The same \(p\) must apply at every trial, under the stated model. A changing environment or a changing definition of success may change the probability.
- Are trials independent? Learning one trial’s result must not alter the probability mechanism for another trial. Dependence can arise from shared conditions or sampling without replacement.
The identical-distribution condition and the fixed-\(p\) condition are closely related here: each indicator must have the same Bernoulli distribution. Independence is a separate requirement. A sequence can have the same success probability on every trial and still be dependent; conversely, independent trials can have different success probabilities.
The distinction from a binomial model is the stopping rule. A binomial variable counts successes in a prespecified number of trials. A waiting-time variable counts trials until a prespecified success occurs. For the latter, the number of trials is not fixed in advance; it is determined by the observed outcomes.
Why the Product of Trial Probabilities Matters
The role of independence can be seen by considering a particular ordered sequence of outcomes. Suppose the first \(k-1\) trials are failures and trial \(k\) is a success. Under independent trials with common success probability \(p\), the probability of this sequence is the product of its trial probabilities:
There are \(k-1\) failures, each with probability \(1-p\), followed by one success with probability \(p\). Independence permits multiplication, and the common probability makes the same factor apply at each position. This is why a geometric model can describe the first-success waiting time. If trial probabilities change, or outcomes are dependent, this product generally changes and the geometric model no longer follows from the assumptions.
The same reasoning applies when waiting for a fixed \(r\)th success: the stopping rule is the position of that success. The geometric case is the special case \(r=1\). A negative binomial model is not justified merely because a question asks for the third or fourth occurrence; the Bernoulli trial assumptions still have to hold.
Worked Example: First Flagged Record in an Independent Review
Hypothetical setting. An insurer’s review team examines policy records one at a time. Suppose each record independently has probability \(0.08\) of being flagged, and the team continues until it finds the first flagged record. Find the probability that the first flagged record is the fourth one reviewed.
Assumption check. A trial is the review of one record, and success means “the record is flagged.” The problem states that the probability is the same for each record and that the outcomes are independent. Thus these are Bernoulli trials with \(p=0.08\), and the first-success waiting time is geometric. This is a hypothetical statistical model, not a claim about actual insurer records. Interest is not applicable, and no mortality assumption is involved.
Calculation. For the first flagged record to be number four, records one, two, and three must all be unflagged, and record four must be flagged. The failure probability is \(1-p=1-0.08=0.92\). Therefore:
Rounded to four decimal places, the probability is \(0.0623\). The calculation checks: it requires exactly three initial failures followed by a success, not merely at least one flagged record among four.
Method and mistake avoided. Identify the first-success stopping rule, then multiply the probabilities of the required ordered outcomes. A common error is to use a binomial probability for exactly one flagged record in four reviews; that event allows the flagged record to occur in any of the four positions, while this question requires it to occur in the fourth position.
Waiting for More Than One Success
If the stopping rule is the \(r\)th success for a fixed \(r>1\), the candidate model is negative binomial. The trials must still have a common success probability and be independent. In an exact-time calculation, the final trial must be a success, and the earlier trials must contain the other \(r-1\) successes. The arrangement of those earlier outcomes matters.
For a particular ordered sequence with \(r\) successes and \(n-r\) failures, independence gives probability \(p^r(1-p)^{n-r}\). If the question specifies that the \(r\)th success occurs on trial \(n\), the final trial is already known to be a success; the remaining \(r-1\) successes can occupy different positions among the first \(n-1\) trials. This explains why counting possible arrangements is needed in addition to multiplying the probability of one sequence.
Worked Example: Third Acceptable Component
Hypothetical setting. A quality analyst tests components one at a time. Each component independently meets a specified acceptance criterion with probability \(0.20\). The analyst stops at the third acceptable component. Find the probability that this third success occurs on test five.
Assumption check. One component test is one Bernoulli trial; “acceptable” is success; \(p=0.20\) is fixed; and outcomes are independent. The target is the third success, so the waiting-time model is negative binomial with parameters \(r=3\) and \(p=0.20\). The scenario is hypothetical. Interest is not applicable, and no mortality assumption is involved.
Calculation. Test five must be a success. Among tests one through four, exactly two must be successes and two failures; otherwise the third success would not occur on test five. There are \(\binom{4}{2}=6\) ways to choose the two success positions in the first four tests. Each such sequence has probability:
The six possible sequences are disjoint, so:
Rounded to four decimal places, the probability is \(0.0307\). As a check, every counted sequence has exactly three acceptable components in the first five tests, with the fifth test providing the third.
Method and mistake avoided. Count the allowed arrangements before multiplying by the probability of one arrangement. Using only \((0.20)^3(0.80)^2\) omits five other valid arrangements for the first four tests.
When the Assumptions Fail
A waiting-time distribution can be the wrong model even when the setting sounds like repeated trials. In a sequence with changing probabilities, the chance of success on the next trial may depend on its position. With dependent outcomes, information about earlier trials may change the chance of success. In either case, a geometric or negative binomial model with one fixed parameter \(p\) is not justified by the stated process.
Worked Example: A Changing Daily Alert Probability
Hypothetical setting. A monitoring system is checked on three successive days. Suppose the daily alert probabilities are \(0.10\), \(0.20\), and \(0.30\), respectively, and the daily outcomes are independent. Find the probability that the first alert occurs on day three, and decide whether a geometric model with one fixed parameter applies.
Assumption check. A trial is a daily check and an alert is success. The outcomes are independent, but the success probabilities differ by day. The trials are not identically distributed, so a geometric model with a single \(p\) does not apply. The setting is hypothetical. Interest is not applicable, and no mortality assumption is involved.
Calculation. No alert on day one has probability \(1-0.10=0.90\). No alert on day two has probability \(1-0.20=0.80\). An alert on day three has probability \(0.30\). By independence:
The result is \(0.2160\), or \(21.60\%\). It is computed directly from the day-specific probabilities, not from a geometric distribution. The check is that the event requires no alert on each of the first two days and an alert on the third.
Method and mistake avoided. Use the stated probability for each trial. Assigning a single \(p\), such as the average \((0.10+0.20+0.30)/3=0.20\), would incorrectly treat each day as having the same success probability.
Worked Example: Inspection Without Replacement
Hypothetical setting. A sealed box contains eight components, of which three are defective and five are acceptable. Components are inspected in random order without replacement. Find the probability that the first defective component is inspected third, and decide whether the geometric model applies.
Assumption check. Define a defective component as a success. The probability of a defect on the first inspection is \(3/8\), but after an acceptable component is removed it becomes \(3/7\), then \(3/6\). The trials are dependent and their conditional probabilities change. Therefore, a geometric model with fixed \(p\) does not apply. This is a hypothetical inspection setting. Interest is not applicable, and no mortality assumption is involved.
Calculation. For the first defect to appear on inspection three, the first two components must be acceptable, followed by a defective component. The successive conditional probabilities are:
The probability is \(0.1786\) to four decimal places. The factors reflect the changing contents of the box: after two acceptable components are removed, three defective components remain among six components. This changing conditional probability confirms why the fixed-\(p\) geometric assumption fails.
Method and mistake avoided. Update the composition after each draw. Treating every inspection as an independent trial with \(p=3/8\) would ignore sampling without replacement.
Boundary Cases and a Common Exam Trap
The probability parameter has meaningful boundary cases. At \(p=1\), all trials succeed, so the waiting time to the first success is \(1\), and the waiting time to the \(r\)th success is \(r\). These are degenerate distributions, not uncertain waiting times. At \(p=0\), no success occurs; the waiting time is infinite and is not a standard finite-valued geometric or negative binomial random variable.
A frequent trap is to name a waiting-time distribution from the wording alone, without checking how the trial process works. “Until the first defect” sounds geometric, but sampling from a fixed batch without replacement violates independence and fixed probability. Another trap is to overlook that a fixed number of successes must be specified: stopping at the third success supports a negative binomial model under the assumptions, while stopping at a random, outcome-dependent target may require a different analysis.
Check Your Understanding
For each problem, first check the trial assumptions and then identify or calculate the requested waiting-time result.
- A hypothetical file review has independent trials, with each file flagged with probability \(0.12\). The review stops at the first flagged file. Name the waiting-time model and its parameter. (Answer: Geometric, \(p=0.12\).)
- Independent sensor checks each detect a fault with probability \(0.25\). What is the probability that the first detection occurs on check three? (Answer: \(0.75^2(0.25)=0.140625\), or \(0.1406\) to four decimals.)
- Independent laboratory tests each pass with probability \(0.40\). What is the probability that the second pass occurs on test four? (Answer: \(\binom{3}{1}(0.40)^2(0.60)^2=0.1728\).)
- A process has independent trials, each with success probability \(1\), and stops at the fourth success. State the waiting time. What changes if the success probability is zero instead? (Answer: With \(p=1\), the waiting time is \(4\) with probability \(1\); with \(p=0\), the waiting time is infinite and is not a standard finite-valued negative binomial variable.)
- A hypothetical system has independent daily success probabilities \(0.15\) on day one and \(0.35\) on day two. Find the probability that its first success is on day two, and say whether one geometric parameter applies. (Answer: \((1-0.15)(0.35)=0.2975\); no single-\(p\) geometric model applies because the probabilities differ.)