The Negative Binomial Failure-Count Model

Stopping at a Target Number of Successes

Some counts arise from observing a fixed number of opportunities, as in the binomial model. Other counts arise by continuing trials until a specified event has happened a fixed number of times. For example, a hypothetical inspection process might continue until four items pass. The number of failures observed before the fourth pass is not fixed in advance; it depends on the trial outcomes.

The negative binomial failure-count model describes this stopping rule. Its parameters have a direct interpretation: one gives the probability of success on each trial, and the other gives the number of successes required before observation stops. Keeping the target number of successes separate from the number of failures is essential when translating a scenario into the model.

Definition: Suppose trials are independent, each trial has the same probability \(p\) of success, and observation stops when the \(r\)th success occurs, where \(r\) is a fixed positive integer. Let \(K\) be the number of failures observed before that \(r\)th success. Then \(K\) has a negative binomial failure-count distribution with parameters \(r\) and \(p\), where \(0<p<1\). Each trial has failure probability \(1-p\).

The word failure is a label for the outcome that is not designated as a success in the model. It does not necessarily mean a defective product or an undesirable event. In an application, a success could mean a claim occurring, a test passing, or a payment being approved; the analyst must define the two outcomes from the scenario.

The stopping rule gives an immediate relationship between the total number of trials and the failure count. If the \(r\)th success occurs on trial \(n\), then the first \(n-1\) trials contain \(r-1\) successes and \(n-r\) failures. The last, \(n\)th trial must be a success. Equivalently, if there are \(k\) failures before stopping, the stopping trial is \(k+r\).

\text{Total trials through the }r\text{th success}=K+r

Probability of a Specified Failure Count

To find the probability that \(K=k\), consider a sequence ending at the \(r\)th success. The final trial is forced to be a success because it is the trial on which the target is reached. In the \(k+r-1\) trials before it, there must be exactly \(r-1\) successes and \(k\) failures. The number of ways to arrange those outcomes is \(\binom{k+r-1}{r-1}\). Each such arrangement has probability \(p^{r-1}(1-p)^k\), and the final success contributes another factor of \(p\).

Multiplying the number of allowable sequences by the probability of each sequence gives the probability function. This argument also explains why the final trial cannot be counted as an unrestricted position: doing so would include sequences that do not stop on that trial.

P(K=k)=\binom{k+r-1}{r-1}p^r(1-p)^k,\qquad k=0,1,2,\ldots

Support: \(K\) can be any nonnegative integer. It may be zero if the first \(r\) trials are all successes, or it may be arbitrarily large if many failures occur before the target number of successes. The parameter \(r\) is the required number of successes; \(p\) is the probability of success on each individual trial.

This model requires a sequence of independent Bernoulli trials with a constant success probability. It is not appropriate if the success probability changes from trial to trial, if outcomes are dependent, or if observation stops for a reason other than reaching the specified number of successes. Those assumptions describe the trial process, not merely the appearance of the observed data.

Worked Example: Third Calibration Pass on Trial Seven

A hypothetical calibration team repeats the same test until a device has passed three times. Each test result is independent, with probability \(0.6\) of a pass. Find the number of failures if the third pass occurs on test seven, and calculate the probability of that stopping time.

Method and parameters. A pass is a success, so \(p=0.6\), and the target is \(r=3\) successes. If the third success occurs on trial \(7\), the number of failures before it is \(k=7-3=4\).

The final test must be a pass. Among the first six tests, there must therefore be two passes and four failures. There are \(\binom{6}{2}=15\) ways to place the two passes in those six positions. Each sequence, including the final pass, has probability \((0.6)^3(0.4)^4\).

P(K=4)=\binom{6}{2}(0.6)^3(0.4)^4 =15(0.216)(0.0256) =15(0.0055296) =0.082944

Thus there are 4 failures, and the probability that the third pass occurs on test seven is 0.0829 to four decimal places. The check is consistent with the stopping rule: three passes plus four failures account for all seven tests. This method avoids treating all seven test positions as freely arranged; the seventh outcome is necessarily a pass.

Worked Example: Fourth Pass on Test Nine

A hypothetical repair team repeats an independent test with probability \(0.8\) of passing each time and stops at the fourth pass. If the fourth pass occurs on test nine, find the number of failures and the probability of this event.

Method and parameters. Here \(p=0.8\), \(r=4\), and the stopping trial is \(9\). Therefore \(k=9-4=5\) failures occurred before the fourth pass. The first eight tests must contain exactly three passes and five failures; the ninth test is the fourth pass.

P(K=5)=\binom{8}{3}(0.8)^4(0.2)^5 =56(0.4096)(0.00032) =56(0.000131072) =0.007340032

The probability is 0.0073 to four decimal places. As a check on the sequence count, \(\binom{8}{3}=\binom{8}{5}=56\): choosing the three pass positions among the first eight is equivalent to choosing the five failure positions.

A common incorrect calculation is \(\binom{9}{4}(0.8)^4(0.2)^5\). It treats the four pass positions as though they could be chosen anywhere among all nine tests, even though the ninth test must be a pass. The correction is to choose the first three passes among the first eight tests, giving \(\binom{8}{3}\), and then include the forced final pass in the factor \((0.8)^4\). This distinction materially changes the answer.

Interpreting the Parameters Before Calculating

The parameter \(r\) is not the total number of trials, and \(p\) is not the probability that the process stops on a particular trial. The process stops as soon as the \(r\)th success occurs. The failure count \(K\) is the random quantity that can vary from one repetition of the process to another.

A reliable translation is to identify the success outcome, count how many successes trigger stopping, and then use the stated stopping trial to determine the failures. If stopping occurs on trial \(n\), the required count is \(n-r\). The sequence before that final trial has \(r-1\) successes, not \(r\), because the \(r\)th success is the final outcome.

Define success.
State which outcome is called a success and identify its probability \(p\).

Identify the stopping target.
Record the fixed number \(r\) of successes required before the process ends.

Translate the stopping time.
If the target is reached on trial \(n\), then \(k=n-r\) failures occurred, and the final trial is a success.

Count allowable sequences.
Arrange \(r-1\) successes and \(k\) failures in the first \(k+r-1\) trials, then include the final success.

Worked Example: Second Approval on the Fifth Review

A hypothetical review process examines applications one at a time until two have been approved. Assume independent outcomes with probability \(0.4\) of approval on each review. Find the probability that the process ends on review five.

Method and parameters. Approval is success, so \(p=0.4\), and \(r=2\). If the second approval is on review five, then \(k=5-2=3\) failures occurred. The first four reviews must contain one approval and three non-approvals, followed by an approval on review five.

P(K=3)=\binom{4}{1}(0.4)^2(0.6)^3 =4(0.16)(0.216) =4(0.03456) =0.13824

The probability is 0.1382 to four decimal places. The outcome count checks: one approval and three non-approvals in the first four reviews, followed by the second approval, gives two approvals and three failures over five reviews. The calculation avoids counting sequences in which the target had already been reached before review five.

Common Traps and Model Checks

The most frequent errors are parameter or stopping-time errors rather than difficult arithmetic. Before evaluating a probability, check that the scenario really has a fixed target number of successes, that the stated probability is for the outcome designated as success, and that the final trial is treated as the success that ends observation.

Reversing the count. The model counts failures before the target success, not successes before the target. If the \(r\)th success occurs on trial \(n\), the failure count is \(n-r\).
Using the wrong number of free positions. For \(K=k\), there are \(k+r-1\) unrestricted positions before the final, forced success—not \(k+r\).
Changing the meaning of success. If the question calls a claim occurrence the success, use the probability of a claim, not the probability of no claim.
Ignoring the assumptions. A stopping rule alone does not establish a negative binomial model. Independence and the same success probability on every trial are also required.
Confusing parameterizations. Some settings describe the total number of trials rather than the failures before stopping. Read the random quantity in the question rather than relying on the distribution’s name alone.

Key takeaway: In the negative binomial failure-count model, \(r\) successes are required, each trial succeeds independently with probability \(p\), and \(K\) counts the failures before the \(r\)th success. For \(K=k\), the final trial is a success and the preceding \(k+r-1\) trials contain \(r-1\) successes and \(k\) failures.

Check Your Understanding

For each problem, identify the target success count and translate the stopping rule before calculating.

A hypothetical materials test has independent pass probability \(0.7\). Testing stops at the second pass. How many failures occur if the second pass is on test four, and what is the probability of that stopping time? (Answer: 2 failures; probability \(=0.1323\))
A hypothetical claim-screening process continues until three claims are flagged. Each record is independently flagged with probability \(0.6\). What is the probability that the third flag occurs on record six? (Answer: \(0.1382\))
A hypothetical permit process ends after two approvals. Each application is independently approved with probability \(0.4\). What is the probability that there are no failures before stopping? (Answer: \(0.1600\))
A hypothetical sensor is checked independently until its third successful reading; the success probability is \(0.8\). Find the probability that the third success is on check five. (Answer: \(0.1229\))
A hypothetical inspection program stops after five units pass, with independent pass probability \(0.5\). If the fifth pass occurs on inspection eleven, how many failures occurred, and what is the probability of this event? (Answer: 6 failures; probability \(=0.1025\))

What You'll Learn