Calculating Exact Negative Binomial Probabilities

Introduction

An exact negative binomial probability depends on correctly reading two details before calculating: what counts as a success, and what quantity the random variable counts. In the failure-count convention, the random variable is the number of failures observed before a specified number of successes. If a problem instead gives the total number of trials, that total must be translated into a failure count before using the formula.

This tutorial focuses on calculating a point probability under that convention. The model assumes independent trials with the same success probability. Once the success event, the required number of successes, and the count are clear, the calculation follows from counting the possible trial sequences.

Definition: Suppose independent trials each have success probability $p$, where $0<p<1$. Let $N$ be the number of failures before the $r$th success, where $r$ is a positive integer. Then $N$ has a negative binomial failure-count distribution, with $$ P(N=n)=\binom{n+r-1}{n}(1-p)^n p^r,\qquad n=0,1,2,\ldots. $$ If $p=1$, every trial is a success and $N=0$ with probability 1. The value $p=0$ is excluded: the $r$th success would never occur, so there would be no finite failure count with the stated distribution.

Identify the Convention Before Substituting

In the formula, $r$ is the number of successes required, $p$ is the probability of success on each trial, and $n$ is the number of failures before the $r$th success. The success must be defined in the language of the scenario; it is not automatically a claim. For example, a problem might define success as an inspection passing or a policy period having no claim.

The stopping trial is necessarily a success: it is the $r$th success. Thus, when there are $n$ failures before that trial, the first $n+r-1$ trials contain $n$ failures and $r-1$ successes, and the final trial is a success. If a problem states that the process took $t$ trials to reach the $r$th success, then the failure count is $n=t-r$. Do not substitute $t$ for $n$.

The combination factor is essential. It counts the different possible arrangements of the $n$ failures and $r-1$ earlier successes in the trials before the final success. Each such arrangement has the same probability, $(1-p)^n p^r$, so multiplying by the number of arrangements gives the point probability.

Why the Formula Has This Form

Consider any particular sequence that ends at the $r$th success and contains $n$ failures. It has $r$ successes and $n$ failures, so independence gives the probability $p^r(1-p)^n$ for that sequence. The final trial is fixed as a success. Among the preceding $n+r-1$ trials, choose which $n$ positions are failures; the remaining positions are the $r-1$ earlier successes. There are $\binom{n+r-1}{n}$ such sequences. These possible sequences are mutually exclusive, so their probabilities add: $$ P(N=n)=\binom{n+r-1}{n}p^r(1-p)^n. $$

This argument uses independent trials with a common success probability. If trial probabilities vary or outcomes are dependent, the sequence probabilities need not be equal, and this formula does not apply as written.

Worked Exact Probabilities

Worked Example: Three Rejections Before the Second Pass

A hypothetical quality-control process checks items one at a time. Each item independently passes with probability $0.60$. Let $N$ be the number of rejected items before the second passing item. Find the probability of exactly three rejections.

Method and parameter identification. A pass is a success, so $p=0.60$. The process stops at the second pass, so $r=2$. The requested count is already failures before that success, so $n=3$. The model assumptions are independent checks and a constant pass probability of $0.60$.

There are $\binom{3+2-1}{3}=\binom{4}{3}=4$ possible sequences: the first four checks have one pass and three rejections, followed by the second pass. Each sequence has probability $0.60^2(0.40)^3$.

Therefore, $$ P(N=3)=\binom{4}{3}(0.40)^3(0.60)^2 =4(0.064)(0.36) =4(0.02304) =0.09216. $$

Rounded to four decimal places, the probability is 0.0922. The sequence check agrees: four possible sequences, each with probability $0.02304$, give $4(0.02304)=0.09216$. The method avoids treating the stopping pass as a rejection or omitting it from the factor $p^r$.

Worked Example: Two Claim-Containing Periods Before the Third Claim-Free Period

Suppose, hypothetically, that successive policy periods are independent, and each period is claim-free with probability $0.50$. Let $N$ count the claim-containing periods before the third claim-free period. Find the probability that $N=2$.

Method and parameter identification. The problem defines a claim-free period as a success, so $p=0.50$. The target is the third success, giving $r=3$. The requested two claim-containing periods are failures, so $n=2$. The stipulated independent, identical period outcomes make the negative binomial model applicable.

The first $n+r-1=4$ periods must contain two claim-containing periods and two claim-free periods. Their positions can be chosen in $\binom{4}{2}=6$ ways, and the fifth period is the third claim-free period. Each sequence has probability $(0.50)^2(0.50)^3=(0.50)^5=1/32$.

Thus, $$ P(N=2)=\binom{4}{2}(0.50)^2(0.50)^3 =6\left(\frac{1}{32}\right) =\frac{6}{32} =0.1875. $$

The exact probability is 0.1875. As a check, six eligible sequences each have probability $1/32$, so their total is $6/32$. A common error is to use $\binom{5}{2}$: the final period is fixed as the third claim-free period, so only the first four positions are arranged.

Worked Example: Six Trials to the Third Success

A hypothetical testing sequence consists of independent trials, each successful with probability $0.40$. The sequence stops at the third success. Find the probability that stopping occurs on trial 6.

Method and parameter identification. Here the stated count is total trials, $t=6$, not failures. The number of failures before the third success is $n=t-r=6-3=3$. The parameters are $r=3$ and $p=0.40$, with failure probability $1-p=0.60$.

The first five trials must contain exactly two successes and three failures, and trial 6 must be the third success. There are $\binom{5}{3}=10$ ways to position the three failures among the first five trials. Each resulting sequence has probability $(0.40)^3(0.60)^3$.

Therefore, $$ P(\text{third success on trial 6}) =P(N=3) =\binom{5}{3}(0.40)^3(0.60)^3 =10(0.064)(0.216) =10(0.013824) =0.13824. $$

Rounded to four decimal places, the probability is 0.1382. The result is checked by counting ten eligible sequences, each with probability $0.013824$. The key correction is using $n=3$, not $n=6$, in the failure-count formula.

A Reliable Calculation Process

A short sequence of decisions prevents most errors in exact negative binomial calculations. State the event that counts as success before assigning $p$; otherwise, it is easy to use the failure probability in the success position or reverse the interpretation of the scenario.

Define success.
Write down which outcome is a success and its probability $p$. Confirm that each trial is independent and has the same success probability.

Identify the stopping target.
Set $r$ equal to the required number of successes, not the number of failures or the total number of trials.

Translate the requested count.
If the random variable counts failures, use that failure count as $n$. If the problem gives total trials $t$, convert to $n=t-r$.

Substitute and check the sequence structure.
Use $\binom{n+r-1}{n}(1-p)^n p^r$. Confirm that the final trial is a success and that the combination counts only the preceding trials.

Common Errors and Boundary Cases

The formula is compact, but the symbols are easy to misread. The following mistakes change the event being calculated, rather than merely changing the arithmetic.

Using total trials as the failure count. In the six-trial example, incorrectly substituting $n=6$ gives $\binom{8}{6}(0.60)^6(0.40)^3=0.083607552$. The event asks for the third success on trial 6, which means $n=6-3=3$; the correct probability is $0.13824$.
Reversing success and failure probabilities. The factor $p^r$ belongs to the $r$ successes. If success is defined as a claim-free period, then $p$ is the claim-free probability, even if the problem discusses claims.
Putting the final success among the arrangements. The final trial is already specified as the $r$th success. The combination is $\binom{n+r-1}{n}$, not $\binom{n+r}{n}$.
Allowing an invalid boundary value. For $p=1$, no failures occur and the distribution is concentrated at $N=0$. For $p=0$, the required success never occurs; the finite failure-count distribution is not defined.

For $n=0$, the only possible sequence is $r$ consecutive successes, so the formula gives $P(N=0)=p^r$ when $0<p<1$. This is a useful direct check on the interpretation of the zero-failure event.

Key takeaway: In the negative binomial failure-count convention, $n$ counts failures before the $r$th success. Set $p$ to the probability of the explicitly defined success, translate any total-trial count using $n=t-r$, and include the final success through the factor $p^r$.

Check Your Understanding

For each problem, identify the success event and the failure count before substituting. Probabilities are rounded to four decimal places where needed.

Independent inspections pass with probability $0.80$. Let $N$ be the number of rejections before the second pass. Find $P(N=1)$. (Answer: 0.2560)
Independent policy periods are claim-free with probability $0.50$. Find the probability of exactly two claim-containing periods before the fourth claim-free period. (Answer: 0.1563)
Independent trials succeed with probability $0.30$. Find the probability that the second success occurs on trial 5. (Answer: 0.1235)
Independent trials succeed with probability $0.60$. Find the probability of no failures before the third success. (Answer: 0.2160)
Independent trials succeed with probability $0.40$. Find the probability that the second success occurs on trial 7. (Answer: 0.0746)
Independent trials succeed with probability $0.25$. Find the probability of exactly four failures before the third success. (Answer: 0.0742)

What You'll Learn