Negative Binomial Tail Probabilities

Why Negative Binomial Tails Need Care

A frequency question often asks whether a count reaches a threshold: for example, whether a hypothetical portfolio has at least a certain number of claims during a year. For a negative binomial count, the probability of exactly $n$ claims is available from its probability mass function. A tail probability adds the probabilities of all counts at or above a threshold. Since there are infinitely many possible counts, it is often easier to calculate the tail as one minus a finite cumulative probability.

The main sources of error are not complicated arithmetic. They are using the wrong negative binomial parameterization, confusing “at least” with “more than,” or including the threshold value on the wrong side of the cumulative probability. Establish the count definition and translate the event before calculating.

Model and Parameterization

Definition: In this tutorial, $N$ is the number of failures observed before the $r$th success in a sequence of independent trials. Each trial has success probability $p$, where $r$ is a positive integer and $0<p<1$. Write $q=1-p$. Then $N$ has a negative binomial distribution with probability mass function $$ P(N=n)=\binom{n+r-1}{n}p^r q^n,\qquad n=0,1,2,\ldots. $$ Here $p$ is the success probability, and $N$ counts failures, not total trials.

The support begins at zero because the $r$th success could occur without any failures. The parameterization matters: some problems instead define the random variable as the total number of trials through the $r$th success. If that total is $T$, then $T=N+r$. A model described in words must be matched to the random variable before its probabilities are calculated.

In a claim-frequency application, a negative binomial model may be specified directly for the annual claim count $N$. In that setting, use the model’s stated $r$ and $p$; do not infer a success-trial story that the problem has not supplied. The probabilities below apply only when the count follows the stated negative binomial model and parameterization.

From Point Probabilities to Tail Probabilities

For an integer threshold $k$, “at least $k$” means $N\geq k$, while “more than $k$” means $N>k$, or equivalently $N\geq k+1$. Define the cumulative distribution function by $F_N(k)=P(N\leq k)$. For $k\geq1$, the inclusive upper tail is

P(N\geq k)=1-F_N(k-1) =1-\sum_{n=0}^{k-1}\binom{n+r-1}{n}p^r q^n.

For a strict upper tail, the threshold in the cumulative probability changes:

P(N>k)=1-F_N(k) =1-\sum_{n=0}^{k}\binom{n+r-1}{n}p^r q^n.

The cumulative probability is a finite sum because only the point probabilities through the threshold are needed. An equivalent direct method is to add the point probabilities in the tail. That approach can be convenient if a problem provides a short range of tail values, but the complement method usually avoids carrying out an unbounded sum.

Confirm the random variable.
Check whether the count records failures before the $r$th success or total trials through that success.

Translate the event.
Rewrite “at least $k$” as $N\geq k$, and “more than $k$” as $N\geq k+1$.

Use the complement.
For $N\geq k$, calculate $1-F_N(k-1)$; for $N>k$, calculate $1-F_N(k)$.

Check the endpoint.
Make sure the cumulative sum stops immediately before the first value included in the tail.

Why the Complement Formula Works

The complement rule applies because the events $N\leq k-1$ and $N\geq k$ are disjoint and together include every possible nonnegative integer value of $N$. Thus, their probabilities sum to one. The same reasoning gives the strict-tail version: $N\leq k$ and $N>k$ partition the support.

P(N\geq k) =1-P(N\leq k-1) =1-\sum_{n=0}^{k-1}P(N=n) =1-\sum_{n=0}^{k-1}\binom{n+r-1}{n}p^r q^n.

This argument depends on the count taking integer values. In particular, it explains why the inclusive tail starts with the cumulative probability at $k-1$, rather than at $k$: the value $k$ belongs in the event $N\geq k$.

Worked Example: At Least Three Claims

A hypothetical portfolio has an annual claim count $N$ modeled as the number of failures before the second success, with $r=2$ and success probability $p=0.60$. Find the probability of at least three claims in a year. Give the probability to four decimal places.

Method. Use the complement of the finite cumulative probability through two claims. Here $q=1-p=0.40$, and “at least three” is $N\geq3$, so $P(N\geq3)=1-P(N\leq2)$.

The point probabilities needed for the cumulative probability are

P(N=0)=\binom{1}{0}(0.60)^2(0.40)^0=0.3600,

P(N=1)=\binom{2}{1}(0.60)^2(0.40)^1 =2(0.36)(0.40)=0.2880,

P(N=2)=\binom{3}{2}(0.60)^2(0.40)^2 =3(0.36)(0.16)=0.1728.

Therefore,

P(N\leq2)=0.3600+0.2880+0.1728=0.8208, \qquad P(N\geq3)=1-0.8208=0.1792.

The probability is 0.1792. As a check, the direct tail begins with $P(N=3)$, which is $\binom{4}{3}(0.60)^2(0.40)^3=0.09216$. The remaining tail probabilities are positive, so a total tail of $0.1792$ is consistent. The common mistake is to subtract $F_N(3)$, which excludes the $N=3$ outcome from the tail.

Worked Example: More Than Four Failures

Suppose a hypothetical annual count $N$ follows the negative binomial model with $r=3$, $p=0.40$, and $q=0.60$. Calculate the probability that the count is more than four. Give the result to four decimal places.

Method. “More than four” means $N>4$, so the complement is $N\leq4$. Calculate the five point probabilities from zero through four and subtract their sum from one.

P(N=0)=\binom{2}{0}(0.40)^3(0.60)^0=0.064,

P(N=1)=\binom{3}{1}(0.40)^3(0.60) =3(0.064)(0.60)=0.1152,

P(N=2)=\binom{4}{2}(0.40)^3(0.60)^2 =6(0.064)(0.36)=0.13824,

P(N=3)=\binom{5}{3}(0.40)^3(0.60)^3 =10(0.064)(0.216)=0.13824,

P(N=4)=\binom{6}{4}(0.40)^3(0.60)^4 =15(0.064)(0.1296)=0.124416.

The cumulative probability and its complement are

P(N\leq4)=0.064+0.1152+0.13824+0.13824+0.124416 =0.580096,

P(N>4)=1-0.580096=0.419904\approx 0.4199.

The answer is 0.4199. The endpoint check is that $N=4$ is not part of “more than four,” so it must be included in the cumulative probability being subtracted. Subtracting only $P(N\leq3)$ would incorrectly calculate $P(N\geq4)$.

Worked Example: Converting Total Trials to a Failure Count

A hypothetical inspection process continues until three acceptable units have been found. Each inspection independently finds an acceptable unit with probability $p=0.70$. Let $T$ be the total number of inspections through the third acceptable unit. Find the probability that at least six inspections are required, to four decimal places.

Method. The negative binomial point-probability formula in this tutorial applies to the number of failures before the third success, not directly to $T$. Write $N=T-3$. Then $T\geq6$ is equivalent to $N\geq3$. With $r=3$, $p=0.70$, and $q=0.30$, calculate $1-P(N\leq2)$.

P(N=0)=\binom{2}{0}(0.70)^3=0.343,

P(N=1)=\binom{3}{1}(0.70)^3(0.30) =3(0.343)(0.30)=0.3087,

P(N=2)=\binom{4}{2}(0.70)^3(0.30)^2 =6(0.343)(0.09)=0.18522.

Thus,

P(N\leq2)=0.343+0.3087+0.18522=0.83692,

P(T\geq6)=P(N\geq3)=1-0.83692=0.16308\approx0.1631.

The probability is 0.1631. The common mistake is to use the threshold six as though it were a failure count. Since $T=N+3$, six total inspections correspond to three failures before the third success, not six failures.

Common Exam Traps

The event wording and the definition of the random variable should be checked separately. A correct point-probability calculation can still produce the wrong answer if either one is misread.

Off by one: For $N\geq k$, subtract $F_N(k-1)$. For $N>k$, subtract $F_N(k)$. The difference is the probability $P(N=k)$.
Reversing $p$ and $q$: In this parameterization, $p$ is the success probability and $q=1-p$ is the failure probability. The formula has $p^r q^n$, because there are $r$ successes and $n$ failures.
Using the wrong count: If the random variable is total trials through the $r$th success, convert with $T=N+r$ before using a failure-count formula.
Stopping the sum too early: For the complement of $N\geq k$, include every value from zero through $k-1$, not just values below $k-1$.
Rounding intermediate terms: Retain full precision in the point probabilities and cumulative sum, then round the final probability as requested.

These checks matter because tail probabilities are used to assess the chance that a count exceeds a practical threshold. The complement method is particularly effective when the threshold is moderate: it replaces an infinite sum with a manageable set of point probabilities and makes the event boundary explicit.

Key takeaway: First identify what the negative binomial count represents. For failures $N$ before the $r$th success, use $P(N=n)=\binom{n+r-1}{n}p^r(1-p)^n$. Then translate the wording exactly: $P(N\geq k)=1-F_N(k-1)$, while $P(N>k)=1-F_N(k)$.

Check Your Understanding

For each problem, identify the event threshold before using the negative binomial probabilities.

A count $N$ is the number of failures before the second success, with $p=0.50$. Find $P(N\geq2)$. (Answer: 0.5000)
A hypothetical annual count follows the failure-count model with $r=3$ and $p=0.60$. Find the probability of more than two events. (Answer: 0.3174)
Let $N$ count failures before the second success, with $p=0.40$. Find the probability of at least four failures. (Answer: 0.3370)
An inspection process has independent success probability $0.80$ and stops at the second success. If $T$ is the total number of trials, find $P(T\geq4)$. (Answer: 0.1040)
For a negative binomial failure count with $r=4$ and $p=0.50$, find the probability of more than three failures. (Answer: 0.5000)

What You'll Learn