Start With the Stopping Rule
Geometric and negative binomial models describe waiting for successes in repeated trials. Their formulas are closely related, so a calculation can look plausible even when the model answers the wrong question. The key decision is not simply how many successes appear in the story: it is which success ends observation.
If the process stops at the first success, the waiting time has a geometric distribution. If it stops at the \(r\)th success, the waiting time has a negative binomial distribution. In either case, first verify the trial assumptions: each trial has two outcomes, success or failure; the trials are independent; and the probability of success is the same value \(p\) on every trial. Write \(q=1-p\).
These definitions count trials, not failures. A different but also standard negative binomial variable counts failures before the \(r\)th success. If \(K_r\) is that failure count, then \(K_r=T_r-r\), so its support is \(k=0,1,\ldots\), and
Always state which variable you are using. For example, “the third success occurs on trial 7” means \(T_3=7\), whereas “there are four failures before the third success” means \(K_3=4\). These are equivalent events because \(7-3=4\), but the values and supports are not interchangeable.
How to Choose the Model
Translate the stopping rule into a sentence before selecting a formula. “Continue until the first accepted item” specifies a geometric waiting time. “Continue until the fourth accepted item” specifies a negative binomial waiting time with \(r=4\). By contrast, “inspect exactly 12 items and count the accepted ones” specifies a fixed number of trials; the count is binomial, not a waiting time. The observation rule and the random variable asked for both matter.
State precisely which outcome of one trial counts as a success, and identify its probability \(p\).
Does observation end at the first success, at the \(r\)th success, or after a fixed number of trials?
For a waiting time, say whether it counts total trials or failures before the target success.
Confirm independent trials with constant \(p\), then verify that the requested value is possible for the chosen variable.
Why the Negative Binomial Formula Has This Form
Suppose observation ends exactly on trial \(t\) at the \(r\)th success. The final trial must be a success. Among the first \(t-1\) trials there must be exactly \(r-1\) successes and \(t-r\) failures. There are \(\binom{t-1}{r-1}\) ways to place those earlier successes. Each such sequence, followed by a final success, has probability \(p^{r-1}q^{t-r}p=p^r q^{t-r}\). The possible sequences are disjoint, so adding their probabilities gives the stated mass function.
Setting \(r=1\) in this result gives \(\binom{t-1}{0}p q^{t-1}=p q^{t-1}\), which is the geometric probability function. Thus the geometric model is the one-success case of the negative binomial model. That relationship does not remove the need to identify the stopping rule: the geometric interpretation is specifically that the first success ends observation.
Worked Example: A Sensor Calibration That Stops at the First Pass
A hypothetical sensor calibration attempt passes with probability \(p=0.4\). Attempts are independent and have the same pass probability. The technician stops as soon as one attempt passes. Let \(T\) be the number of attempts through the first pass. Find (a) the probability that the first pass occurs on attempt 3 and (b) the probability of a pass by the end of attempt 3.
Method and model. The stopping event is the first pass, so \(T\) has a geometric distribution with parameter \(p=0.4\). Thus \(q=1-p=0.6\), and \(T\) can take the values \(1,2,\ldots\).
(a) For the first pass to occur on attempt 3, attempts 1 and 2 must fail and attempt 3 must pass:
(b) A pass by attempt 3 means \(T\leq 3\). The complementary event is that all three attempts fail:
The values are probabilities, so both lie between 0 and 1. As a check, \(P(T\leq 3)=0.4+0.24+0.144=0.784\), agreeing with the complement calculation. The common mistake to avoid is using \(p^3\) for \(P(T=3)\): that would require three passes, whereas this event requires two failures followed by one pass.
Worked Example: A Review Process That Stops at the Third Approval
In a hypothetical review process, each file is approved with probability \(p=0.3\), independently, with the same probability for every file. The process ends when the third approval is recorded. Let \(T_3\) count all files reviewed, including the file that receives the third approval. Find the probability that the process ends on file 7.
Method and model. The target is the third success, so use a negative binomial model with \(r=3\) and \(p=0.3\). Then \(q=0.7\). For \(T_3=7\), the seventh file must be approved, and exactly two of the first six files must be approved. The remaining four of those six are failures.
The probability is approximately \(0.0972\) to four decimal places. The support check also works: a third approval cannot occur before file 3, and file 7 is possible. A common mistake is to use the geometric probability \(q^6p\), which describes the first approval occurring on file 7. Here the process continues after the first and second approvals, so the target is the third approval and the earlier six files must contain two approvals, not zero.
Translate Carefully Between Trials and Failures
A phrase such as “four failures before the second success” asks for a failure count, while “the second success occurs on trial six” asks for a total-trials count. For fixed \(r\), the conversion is \(T_r=K_r+r\). In the failure-count formula, \(k\) failures require \(k+r\) total trials; the final trial is a success, and the other \(r-1\) successes are placed among the preceding \(k+r-1\) trials.
The number of failures is not the number of trials. In particular, when \(r=2\) and there are three failures before completion, the total number of trials is \(3+2=5\), not 3. The supports make this distinction visible: \(T_r\geq r\), while \(K_r\geq 0\).
Worked Example: The Second Success After Three Failures
A hypothetical quality screen classifies each item as acceptable with probability \(p=0.6\), independently with constant \(p\). Screening stops at the second acceptable item. Let \(K_2\) count the failures before that second success. Find the probability of exactly three failures before stopping, and state the equivalent total-trials event.
Method and model. This is a negative binomial failure count with \(r=2\), \(p=0.6\), and \(q=0.4\). The event \(K_2=3\) is equivalent to \(T_2=3+2=5\): the second success occurs on trial 5.
Among the first four trials there must be exactly one success and three failures. Trial 5 must be the second success. Therefore:
Thus the probability is \(0.0922\) to four decimal places. As a check, the total-trials formula gives \(\binom{5-1}{2-1}(0.6)^2(0.4)^{5-2}=4(0.36)(0.064)=0.09216\), the same result. The common mistake is to call this \(T_2=3\); three is the number of failures, and the correct total is five trials.
When Neither Waiting-Time Model Applies
Geometric and negative binomial models require the process to continue until a specified success occurs. If observation instead ends after a predetermined number \(n\) of trials, the number of successes is a binomial count. For example, inspecting exactly eight items and counting approvals is not a negative binomial experiment just because the approvals are successes. There is no success that triggers the end of observation.
The standard models also require independent trials with a fixed success probability. If success probabilities change from trial to trial, or if outcomes are dependent, the geometric or negative binomial probability functions generally do not apply as written. A without-replacement inspection from a small finite group, for instance, does not have independent trials with a constant \(p\). The stopping-rule language alone is not enough; the probability assumptions must also fit.
Common Exam Traps
- Using the first-success model when the process continues. If the stated target is the \(r\)th success for \(r>1\), a geometric waiting time does not describe the total process.
- Mixing the two negative binomial conventions. A count of failures \(k\) and a count of trials \(t\) satisfy \(t=k+r\); they do not have the same support or numerical value.
- Ignoring the final success. For completion exactly on trial \(t\), trial \(t\) must be a success. The successes among the first \(t-1\) trials total \(r-1\), not \(r\).
- Calling a fixed-horizon count a waiting time. If the number of trials is set in advance, the random variable is the number of successes in those trials, not the number of trials until a target success.
- Assuming the model from a familiar word alone. “Success” is just a label for the outcome of interest. Define it in context and check independence and constant \(p\) before applying a formula.
Check Your Understanding
For each problem, identify the relevant model or stopping rule and use the stated variable convention.
- A hypothetical device passes a test independently with constant probability \(0.25\). Testing stops at the first pass. Let \(T\) count tests through that pass. Find \(P(T=4)\). (Answer: geometric with \(p=0.25\); \(0.75^3(0.25)=0.1055\) to four decimals.)
- Each independently processed application is approved with probability \(0.5\), and processing stops at the third approval. Find the probability that the third approval is on application 5. (Answer: negative binomial for \(T_3\); \(\binom{4}{2}(0.5)^3(0.5)^2=0.1875\).)
- A hypothetical filter retains an item with probability \(0.4\), independently with constant probability, and the process stops at the fourth retained item. Let \(K_4\) count discarded items before stopping. Find \(P(K_4=2)\). (Answer: \(\binom{5}{3}(0.4)^4(0.6)^2=0.09216\), or \(0.0922\) to four decimals.)
- Exactly eight independent trials are conducted, each with success probability \(0.4\). The experiment does not stop early. Identify the model for the number of successes and find the probability of exactly three successes. (Answer: binomial, not geometric or negative binomial; \(\binom{8}{3}(0.4)^3(0.6)^5=0.2787\) to four decimals.)
- A process continues until the second success, with independent trials and constant success probability \(0.7\). Let \(T_2\) count total trials. Find the probability that more than two trials are needed. (Answer: negative binomial stopping rule; \(P(T_2>2)=1-(0.7)^2=0.5100\).)