Interpreting a Negative Binomial Count
A negative binomial count can describe the number of unsuccessful attempts made before a process reaches a specified number of successes. In a hypothetical claims operation, for example, an employee might review files one at a time until a fixed number meet a validation standard. The count of files that fail validation before the target number pass is not the same as the total number of files reviewed: the target success is included in the total, but not in the failure count.
The model is built from repeated Bernoulli trials. Each trial has one of two outcomes, success or failure. Here, “success” is simply the outcome named as success for the problem; it does not have to mean a favourable result in an everyday sense. The process stops when the \(r\)th success occurs, where \(r\) is a fixed positive integer.
The assumptions matter. The success probability must be the same on every trial, and outcomes on different trials must be independent. The stopping target \(r\) is fixed in advance. If the probability changes from trial to trial, or if the trials influence one another, this probability function does not describe the process as stated. Also, the model does not apply to a fixed number of trials with a random number of successes; that is a different counting setup.
Why the Probability Function Has This Form
To have exactly \(k\) failures before the \(r\)th success, the final trial must be a success. Before that final trial there must be \(r-1\) successes and \(k\) failures, in any order. This gives a direct way to count the possible histories that end at the target success.
If there are \(k\) failures before the \(r\)th success, the process ends after \(k+r\) trials. The final trial is the \(r\)th success.
In the first \(k+r-1\) trials, there are \(r-1\) successes and \(k\) failures. Choosing which \(r-1\) positions are successes gives \(\binom{k+r-1}{r-1}\) possible arrangements.
Every such history has \(r\) successes and \(k\) failures. By independence, its probability is \(p^r q^k\).
The arrangements are mutually exclusive and have the same probability, so multiply their count by \(p^r q^k\).
Thus the probability function follows:
The coefficient can equivalently be written \(\binom{k+r-1}{k}\): it counts the ways to place the \(k\) failures among the first \(k+r-1\) trials. The last trial is not a choice; it must be the success that reaches the target. When \(k=0\), all \(r\) trials are successes, and the formula gives \(P(K=0)=p^r\), as expected.
The support is \(0,1,2,\ldots\), because there may be no failures before the target success or any nonnegative number of failures. In problems, check carefully what the random variable counts. If a question instead asks for the total number of trials \(T\) up to the \(r\)th success, then \(T=K+r\), so the event \(T=n\) corresponds to \(K=n-r\), for \(n=r,r+1,\ldots\).
Worked Example: Two Rejected Files Before Three Accepted Files
A hypothetical reviewer examines files in sequence. Each file independently passes a specified validation check with probability \(0.60\). The reviewer stops when three files have passed. Find the probability that exactly two files fail before the third file passes. Name the number of possible trial histories.
Method and setup. Let success mean that a file passes. Here \(r=3\), \(p=0.60\), \(q=1-p=0.40\), and the requested failure count is \(k=2\). The process ends after \(k+r=5\) reviews. Its last review must be the third pass.
Count the histories. Among the first four reviews there must be two passes and two failures. Choose the two positions for the passes in \(\binom{4}{2}=6\) ways. Each history has three passes and two failures, so its probability is \(0.60^3(0.40)^2\).
Calculate.
The probability, rounded only at the final step to four decimal places, is 0.2074. There are six possible histories. Each has the final review as a pass and has two passes and two failures in the preceding four reviews.
Common mistake avoided. Counting arrangements across all five reviews would ignore the stopping rule. Only the first four positions can vary; the fifth must be the success that completes the target of three passes.
Applying the Stopping Rule to Events
A question may ask for a range of failure counts rather than one exact value. Apply the probability function to each allowed value of \(k\), then add the probabilities. The events \(K=0\), \(K=1\), and so on are disjoint: one observed process cannot have two different numbers of failures before the same stopping success.
Be especially precise with phrases such as “no more than,” “at least,” or “the target success occurs on trial \(n\).” If the target success occurs on trial \(n\), there are \(n-r\) failures before it. The target success itself is not a failure and is not included in \(K\).
Worked Example: At Most One Failed Record Before Two Passes
In a hypothetical quality review, records are checked sequentially. Each record independently passes with probability \(0.75\). The review stops at the second passing record. Find the probability of at most one failed record before stopping.
Method and setup. Let \(K\) be the number of failures before the second pass. Then \(r=2\), \(p=0.75\), and \(q=0.25\). “At most one” means \(K=0\) or \(K=1\), so the required probability is \(P(K=0)+P(K=1)\).
Zero failures. Both trials must pass:
One failure. The final trial is a pass. The first two trials contain one pass and one failure, which can occur in \(\binom{2}{1}=2\) orders:
Add the disjoint cases.
Rounded only at the final step to four decimal places, the answer is 0.8438. The method avoids treating “at most one” as exactly one: the zero-failure case must also be included.
Worked Example: The Third Pass on the Fifth Review
A hypothetical assessor reviews applications one at a time. Each application independently meets a specified requirement with probability \(0.40\). The assessor stops at the third application that meets the requirement. Find the probability that the third pass occurs on review five, and explain what the failure count is in this event.
Method and setup. The third pass on review five means there are exactly \(5-3=2\) failures before the third pass. Thus \(r=3\), \(k=2\), \(p=0.40\), and \(q=0.60\). The final review must be a pass.
Count and calculate. The first four reviews contain two passes and two failures. There are \(\binom{4}{2}=6\) arrangements. Each arrangement followed by the required final pass has probability \((0.40)^3(0.60)^2\). Therefore,
Rounded only at the final step to four decimal places, the probability is 0.1382. The failure count is two, not five: five is the total number of reviews, consisting of three passes and two failures. This solution avoids confusing the trial number of the stopping success with the number of failures.
Common Sequence-Counting Trap
A frequent error is to choose positions for the successes among all \(k+r\) trials, using \(\binom{k+r}{r}\). That would be appropriate if the question were about arrangements of \(r\) successes and \(k\) failures in a fixed number of trials without a stopping rule. Here, however, the process stops at the \(r\)th success, so the final trial must be a success.
For the first worked example, using \(\binom{5}{3}=10\) would give the incorrect calculation
The correct coefficient is \(\binom{4}{2}=6\), because only the first four trial positions can vary; the fifth is forced to be the third success. The corrected probability is \(6(0.03456)=0.20736\), or \(0.2074\) to four decimal places. The error counted five-trial sequences that ended in a failure, even though the process would already have stopped at the third success in such a sequence.
Check Your Understanding
For each hypothetical process, identify the failure count and use the stopping rule to choose the relevant negative binomial probability. Probabilities are rounded to four decimal places where needed.
- A sequence of independent equipment checks each passes with probability \(0.40\). The process stops at the first pass. Find the probability of exactly three failures before that pass. (Answer: \(0.0864\))
- A hypothetical auditor checks entries, each independently acceptable with probability \(0.50\), and stops at the second acceptable entry. Find the probability of exactly two unacceptable entries before stopping. (Answer: \(0.1875\))
- In a hypothetical review, each file independently passes with probability \(0.80\), and review stops at the third pass. Find the probability of no more than one failure before stopping. (Answer: \(0.8192\))
- A sequence of applications independently meets a requirement with probability \(0.60\). The assessor stops at the second application that meets it. Find the probability that this second success occurs on the fourth application. (Answer: \(0.1728\))
- A hypothetical test produces independent passes with probability \(0.25\), and testing stops at the fourth pass. Find the probability that there are no failures before stopping. (Answer: \(0.0039\))