Negative Binomial Tail Probabilities By Complements

Why Negative Binomial Tails Need Careful Cutoffs

A question about an exact number of failures asks for one probability. A question about at least or more than a specified number asks for a tail, which can involve many possible counts. In the negative binomial failure-count model, the count has no finite upper bound, so adding every term in an upper tail is generally not the most efficient method. The key is to identify the event boundary precisely and, when convenient, subtract the corresponding lower-tail probability from one.

This tutorial uses the convention that the random variable counts failures before a fixed number of successes. The success probability and the number of required successes determine the point probabilities. Once these are specified, a tail calculation is a matter of translating the wording into the right integer event and evaluating it without dropping or adding a boundary term.

Definition: Let $K$ be the number of failures observed before the $r$th success in a sequence of independent Bernoulli trials. Each trial has success probability $p$, where $0<p<1$, and failure probability $q=1-p$. The parameter $r$ is a positive integer. Then $K$ has a negative binomial failure-count distribution with support $k=0,1,2,\ldots$ and point probabilities $$ P(K=k)=\binom{k+r-1}{k}p^r q^k. $$

The formula counts the ways to arrange the first $k+r-1$ trials with $k$ failures and $r-1$ successes, followed by the $r$th success. Thus the final trial is a success, and the probability of any such sequence is $p^r q^k$. This convention matters: $K$ counts failures, not the total number of trials. The number of trials up to and including the $r$th success is $K+r$.

Translating Tail Wording into Events

Because $K$ takes integer values, “at least $m$” and “more than $m$” have different lower boundaries. Write the event first, before applying a formula. For an integer $m\geq 0$, “at least $m$” includes $m$, while “more than $m$” starts at $m+1$.

Wording	Event for $K$	Complement
At least $m$ failures	$K\geq m$	$K\leq m-1$
More than $m$ failures	$K>m$, equivalently $K\geq m+1$	$K\leq m$

Define the lower-tail cumulative probability at an integer $j\geq 0$ by $F_K(j)=P(K\leq j)$. For convenience, set $F_K(j)=0$ when $j<0$. Then the upper-tail formulas are:

P(K\geq m)=1-F_K(m-1) =\sum_{k=m}^{\infty}P(K=k), \qquad P(K>m)=1-F_K(m) =\sum_{k=m+1}^{\infty}P(K=k).

The two forms of each probability are equivalent. The complement form requires only the point probabilities up to the cutoff. A direct sum can be convenient when only a few upper-tail terms are needed, or when a problem specifically asks for that sum. Since the support is unbounded, a direct upper-tail sum has infinitely many terms; in routine calculations, summing a finite lower tail and taking the complement is often more practical.

Why the Complement Formula Works

The complement method follows from a partition of the possible values of $K$. For “at least $m$,” every nonnegative integer value is either at most $m-1$ or at least $m$, and the two events cannot occur together.

Split the support.
The events $K\leq m-1$ and $K\geq m$ are disjoint and together cover every possible value of $K$.

Add their probabilities.
Since the events form a partition, $P(K\leq m-1)+P(K\geq m)=1$.

Rearrange.
Subtracting the lower-tail probability gives $P(K\geq m)=1-P(K\leq m-1)$. For “more than $m$,” the omitted values are $0,1,\ldots,m$, so $P(K>m)=1-P(K\leq m)$.

There is also a useful recurrence for successive point probabilities. If $b_k=P(K=k)$, then dividing the formula for $b_{k+1}$ by the formula for $b_k$ gives:

\frac{b_{k+1}}{b_k} =q\frac{k+r}{k+1}, \qquad b_{k+1}=b_kq\frac{k+r}{k+1}.

This follows because the ratio of the binomial coefficients is $\binom{k+r}{k+1}/\binom{k+r-1}{k}=(k+r)/(k+1)$, and the power of $q$ gains one factor of $q$. The recurrence is a check on successive terms and can reduce repeated computation. It does not change the cutoff: the event wording still determines which terms belong in the lower tail or upper tail.

Worked Example: At Least Three Failures

A hypothetical claim-count process is modeled as independent trials, each with success probability $p=0.5$. The count $K$ is the number of failures before the second success, so $r=2$. Find the probability of at least 3 failures. Give the answer to four decimal places. Use the lower-tail complement, and identify the boundary term that must be included.

Method. “At least 3” means $K\geq 3$, so the complement is $K\leq 2$. The lower tail must therefore include $k=0,1,2$, not just values below 2.

Here $q=1-p=0.5$, $p^r=(0.5)^2=0.25$, and the point-probability formula becomes $P(K=k)=\binom{k+1}{k}(0.25)(0.5)^k$. Calculate each term:

For $k=0$, $P(K=0)=\binom{1}{0}(0.25)(1)=0.25$. For $k=1$, $P(K=1)=\binom{2}{1}(0.25)(0.5)=2(0.25)(0.5)=0.25$. For $k=2$, $P(K=2)=\binom{3}{2}(0.25)(0.25)=3(0.25)(0.25)=0.1875$.

Thus $P(K\leq 2)=0.25+0.25+0.1875=0.6875$, and

P(K\geq 3)=1-0.6875=0.3125.

Check. The omitted lower-tail probability is $0.6875$, so the remaining probability is $0.3125$; together they total $1$. The answer is $0.3125$. The common mistake this method avoids is excluding $k=2$ from the complement even though $K=2$ is not “at least 3.”

Worked Example: More Than Two Failures

A hypothetical monitoring procedure records failures before its third success. Trials are independent, with success probability $p=0.5$, and $K$ denotes the failure count, so $r=3$. Find the probability of more than 2 failures. Give the answer to four decimal places.

Method. “More than 2” means $K>2$, or equivalently $K\geq 3$. Its complement is $K\leq 2$. This is different from “at least 2,” whose complement would stop at $K\leq 1$.

Here $q=0.5$ and $p^r=(0.5)^3=0.125$. Calculate the terms in the complement:

$P(K=0)=\binom{2}{0}(0.125)(1)=0.125$.

$P(K=1)=\binom{3}{1}(0.125)(0.5)=3(0.125)(0.5)=0.1875$.

$P(K=2)=\binom{4}{2}(0.125)(0.25)=6(0.125)(0.25)=0.1875$.

The lower-tail sum is $P(K\leq 2)=0.125+0.1875+0.1875=0.5$. Therefore:

P(K>2)=1-P(K\leq 2)=1-0.5=0.5000.

Check. The complement includes exactly the three possible values $0,1,2$; the event of interest begins at $3$. The two probabilities sum to $1$. The answer is $0.5000$. The common mistake avoided here is treating “more than 2” as “at least 2,” which would incorrectly include $K=2$.

Using a Complement with Unequal Trial Probabilities

The tail rules do not depend on $p$ being one-half. With another success probability, calculate the lower-tail terms using the specified $p$ and $q=1-p$, then subtract their sum from one. Pay particular attention to powers of $q$: a term for $k$ failures contains $q^k$, while the factor $p^r$ accounts for the $r$ successes.

Worked Example: More Than Three Failures with a Non-half Success Probability

In a hypothetical independent-trial process, success probability is $p=0.4$, and $K$ counts failures before the third success, so $r=3$. Find the probability that there are more than 3 failures. Give the answer to four decimal places.

Method. The event is $K>3$, so its complement is $K\leq 3$. Use $q=1-0.4=0.6$ and $p^r=(0.4)^3=0.064$. For each $k$, apply $P(K=k)=\binom{k+2}{k}(0.064)(0.6)^k$.

For $k=0$, $P(K=0)=1(0.064)(1)=0.064$.

For $k=1$, $P(K=1)=3(0.064)(0.6)=3(0.0384)=0.1152$.

For $k=2$, $P(K=2)=6(0.064)(0.36)=6(0.02304)=0.13824$.

For $k=3$, $P(K=3)=10(0.064)(0.216)=10(0.013824)=0.13824$.

Add the complement terms: $P(K\leq 3)=0.064+0.1152+0.13824+0.13824=0.45568$. Hence:

P(K>3)=1-0.45568=0.54432\approx 0.5443.

Check. The lower-tail probability $0.45568$ plus the desired upper tail $0.54432$ equals $1$. Rounded only at the final step to four decimal places, the answer is $0.5443$. The common mistake avoided is using $p^kq^r$ rather than $p^rq^k$; the model has three successes and $k$ failures in the counted sequence.

Common Boundary Errors and Efficient Checks

Most errors in these problems are not failures of arithmetic; they are failures to translate the wording. A useful discipline is to write an event and its complement in symbols before substituting any numbers. In particular:

“At least $m$” includes $m$. The complement ends at $m-1$: $P(K\geq m)=1-P(K\leq m-1)$.
“More than $m$” excludes $m$. The complement ends at $m$: $P(K>m)=1-P(K\leq m)$.
Use the failure-count support. The possible values start at zero. The count $K$ is not the total trial count, which would be $K+r$.
Keep parameters in their stated roles. In this convention $p$ is the probability of success and $q=1-p$ is the probability of failure. A term for $k$ failures has $q^k$.
Check the complement boundary. List the included lower-tail values when a cutoff is small. For example, $K\leq 2$ contains $0,1,2$, three point probabilities.

For a nonnegative integer threshold of zero, the same rule remains valid: $P(K\geq 0)=1$, since $K$ cannot be negative. The convention $F_K(-1)=0$ makes the formula $1-F_K(m-1)$ work at $m=0$ as well. If a threshold is not an integer, first use the fact that $K$ is integer-valued to restate the event with an integer cutoff; for example, $K\geq 2.5$ is the same event as $K\geq 3$.

A final check is to make sure a calculated probability lies between zero and one and that a lower tail and its complement add to one. If using the recurrence for successive point probabilities, compare at least one term with the original point-probability formula. These checks can reveal a missing boundary term, a wrong power, or a mistaken parameter role before the result is reported.

Key takeaway: Translate the wording first. For at least $m$, subtract $P(K\leq m-1)$ from one; for more than $m$, subtract $P(K\leq m)$ from one. Include every integer in the stated lower-tail complement, and use $p^rq^k$ for the failure-count convention.

Check Your Understanding

For each hypothetical process, use the stated failure-count convention and give the requested probability to four decimal places.

Independent trials have success probability $p=0.5$, and $K$ counts failures before the second success. Find $P(K\geq 2)$. (Answer: 0.5000)
Independent trials have success probability $p=0.5$, and $K$ counts failures before the third success. Find $P(K>1)$. (Answer: 0.6875)
Independent trials have success probability $p=0.6$, and $K$ counts failures before the second success. Find $P(K\geq 3)$. (Answer: 0.1792)
Independent trials have success probability $p=0.4$, and $K$ counts failures before the third success. Find $P(K>3)$. (Answer: 0.5443)
Independent trials have success probability $p=0.25$, and $K$ counts failures before the first success. Find the probability of at least 4 failures. (Answer: 0.3164)

Wording	Event for \(K\)	Complement
At least \(m\) failures	\(K\geq m\)	\(K\leq m-1\)
More than \(m\) failures	\(K>m\), equivalently \(K\geq m+1\)	\(K\leq m\)

What You'll Learn