Two Negative Binomial Supports: Trials and Preceding Failures

Why Two Supports Appear

A negative binomial waiting-time problem begins with repeated Bernoulli trials and ends when a specified number of successes has occurred. The count recorded, however, can be defined in either of two ways: it can include the successful trial that completes the process, or it can count only the failures that came before that final success. These are different random variables, so they have different supports and probability formulas.

The distinction matters whenever a problem asks for an exact waiting time, a possible range of values, or a probability by a deadline. A useful first step is to ask: does the count include the final success trial?

Model and notation: Suppose trials are independent, each trial has success probability $p$, and the same $p$ applies on every trial, where $0<p<1$. Let $r$ be a fixed positive integer, the target number of successes, and let $q=1-p$ be the failure probability. No interest-rate or mortality model is involved in this probability model.

The Two Random Variables and Their Supports

Let $T_r$ denote the number of trials through and including the $r$th success. The process cannot end before $r$ trials, because each trial can produce at most one success. It can end on trial $r$ if the first $r$ trials are all successes, or on any later trial if more failures occur first.

T_r\in\{r,r+1,r+2,\ldots\}.

In this convention, $r$ specifies the number of successes required, while $p$ specifies the success probability on each trial. The value of $T_r$ includes the final successful trial.

Alternatively, let $F_r$ denote the number of failures before the $r$th success. The count can be zero if the first $r$ trials are successes. It can also be any nonnegative integer, depending on how many failures occur before the target is reached.

F_r\in\{0,1,2,\ldots\}.

Here too, $r$ is the required number of successes and $p$ is the per-trial success probability. The final successful trial is not counted as a failure, so it is not included in $F_r$. The supports differ because one variable counts all trials up to completion and the other counts only the failures along the way.

Convention	What is counted?	Support	Smallest possible value
$T_r$	Trials through and including the $r$th success	$r,r+1,r+2,\ldots$	$r$
$F_r$	Failures before the $r$th success	$0,1,2,\ldots$	$0$

Probability Functions for Each Convention

The two probability functions follow by counting the possible arrangements of successes and failures before the final, required success. For $T_r=n$, the $n$th trial must be a success. Among the first $n-1$ trials, there must be exactly $r-1$ successes and $n-r$ failures. For $F_r=k$, the final trial is again a success, while the preceding $r+k-1$ trials contain $r-1$ successes and $k$ failures.

Probability functions: Under the independent, constant-$p$ Bernoulli model, $$ P(T_r=n)=\binom{n-1}{r-1}p^r q^{n-r},\qquad n=r,r+1,\ldots, $$ and $$ P(F_r=k)=\binom{k+r-1}{r-1}p^r q^k,\qquad k=0,1,2,\ldots. $$ Outside the stated support, the probability is zero.

The binomial coefficient in the first expression chooses which $r-1$ of the first $n-1$ trials are successes. In the second expression it chooses which $r-1$ of the $k+r-1$ trials before the final success are successes. The powers $p^r$ and $q$ to the number of failures account for the $r$ successes, including the final one, and the failures, respectively.

This counting argument also explains why the support conditions are built into the formulas. For $T_r$, the expression $n-1$ must leave room for $r-1$ earlier successes, so $n\ge r$. For $F_r$, the number of failures $k$ cannot be negative. These formulas rely on independent trials with a fixed success probability; changing success probabilities or dependence between trials requires a different model.

Worked Examples

Worked Example: Counting Trials Through the Third Success

A hypothetical screening process has independent trials. Each trial gives a positive result with probability $p=0.4$, and the process stops at the third positive result. Let $T_3$ be the number of trials through that third positive result. State the support and calculate the probability that the process ends on trial 5. No interest or mortality model is involved.

Method and model. This is the trials-through-completion convention, with $r=3$, $p=0.4$, and $q=1-p=0.6$. The final trial must be positive. For the process to end on trial 5, exactly two of the first four trials must be positive and the other two must be negative.

Support. At least three trials are needed for three positive results, so $T_3\in\{3,4,5,\ldots\}$.

Calculate. Choose the two positive results among the first four trials, then multiply by the probability of the required three positives and two negatives:

\begin{aligned} P(T_3=5) &=\binom{4}{2}(0.4)^3(0.6)^2\\ &=6(0.064)(0.36)\\ &=6(0.02304)\\ &=0.13824. \end{aligned}

Rounded to four decimal places, $P(T_3=5)=0.1382$. The calculation has three factors of $0.4$, including the positive result on trial 5, and two factors of $0.6$ for the failures. The common mistake this method avoids is using $\binom{5}{3}$, which would allow the final success to be placed before trial 5 even though the process must end with its third success on trial 5. A check of the event structure confirms that trial 5 is positive and the first four contain exactly two positives.

Worked Example: Counting Failures Before the Fourth Success

A hypothetical quality check has independent trials, each accepted with probability $p=0.7$. Inspection stops at the fourth accepted item. Let $F_4$ count the rejected items before the fourth acceptance. State its support and find the probability of exactly two rejections before completion. No interest or mortality model is involved.

Method and model. This is the failures-before-completion convention, with $r=4$, $p=0.7$, and $q=0.3$. The support is $F_4\in\{0,1,2,\ldots\}$, since zero rejections are possible and the number of rejections has no finite upper bound.

Arrange the trials. If there are exactly two rejections before the fourth acceptance, the last trial is an acceptance. Before it, there are $4-1=3$ acceptances and two rejections, making five trials. The three acceptances can occupy any three of those five positions.

\begin{aligned} P(F_4=2) &=\binom{5}{3}(0.7)^4(0.3)^2\\ &=10(0.2401)(0.09)\\ &=10(0.021609)\\ &=0.21609. \end{aligned}

Rounded to four decimal places, $P(F_4=2)=0.2161$. The four factors of $0.7$ represent all four acceptances, including the last one; the two factors of $0.3$ represent the rejections. A common mistake is to use a support starting at 4 for $F_4$. That lower bound belongs to the total trial count, not the failure count. Here zero rejections is possible, which confirms the stated support.

Worked Example: Completing by a Trial Deadline

A hypothetical process has independent trials with success probability $p=0.5$ and stops when its fourth success occurs. Let $T_4$ be the number of trials through that success. Find the probability that the process completes by trial 6. No interest or mortality model is involved.

Method and model. The variable has support $T_4\in\{4,5,6,\ldots\}$. Completion by trial 6 means that, on the underlying sequence of potential trials, at least four of the first six trials are successes. A hypothetical continuation of six trials is useful for defining the event, even though the stopped process would not actually observe trials after its fourth success.

Count the deadline event. In the underlying six-trial sequence, there can be four, five, or six successes. Each specific sequence has probability $(0.5)^6=1/64$, and the numbers of sequences in these cases are $\binom{6}{4}=15$, $\binom{6}{5}=6$, and $\binom{6}{6}=1$.

\begin{aligned} P(T_4\le 6) &=\left[\binom{6}{4}+\binom{6}{5}+\binom{6}{6}\right](0.5)^6\\ &=(15+6+1)\frac{1}{64}\\ &=\frac{22}{64}\\ &=0.34375. \end{aligned}

Rounded to four decimal places, $P(T_4\le6)=0.3438$. The common trap is to calculate only the probability of exactly four successes in six potential trials, $15/64$. That omits underlying sequences with five or six successes, each of which also has the fourth success by trial 6. In the stopped process, more than four successes are not observed: the process ends at the fourth. The extra successes occur only in the hypothetical continuation, and their presence is a valid way to identify sequences whose stopping time is at most 6. The final event is therefore checked as “at least four successes by the deadline,” not “exactly four successes in six trials.”

Reading the Wording and Avoiding Indexing Errors

Problems often describe the count in words rather than name the random variable. The stopping rule is the same in both conventions, but the recorded number changes according to what the wording counts. “Total number of trials until the fourth success, including the successful trial” indicates $T_4$. “Number of failures before the fourth success” indicates $F_4$. If the problem reports the trial on which the target success occurs, it is describing the trials-through-success count.

A useful check is to test the smallest possible value. The trial count cannot be less than $r$; the failure count can be zero. For example, a process that needs four successes cannot finish on its third trial, but it can have zero failures if its first four trials are all successes. This support check often exposes a mistaken interpretation before any probability is calculated.

Key takeaway: Identify exactly what the random variable counts before selecting its support. $T_r$ counts trials, including the final success, and starts at $r$. $F_r$ counts failures only and starts at zero. Both use the same target $r$ and success probability $p$, but their values and probability functions are indexed differently.

Check Your Understanding

For each problem, first identify whether the count is trials through success or failures before success. Answers are rounded to four decimal places where needed.

A process requires two successes, with independent trials and success probability $0.35$. State the support of the trial count and calculate the probability it ends on trial 4. (Answer: support $\{2,3,4,\ldots\}$; $P(T_2=4)=0.1553$)
A device passes independent tests with probability $0.6$. Testing stops at the third pass. State the support for the number of failures and calculate the probability of exactly one failure before stopping. (Answer: support $\{0,1,2,\ldots\}$; $P(F_3=1)=0.2592$)
An independent trial succeeds with probability $0.8$, and the process stops at the fifth success. What is the support of the failure count, and what is the probability of no failures before stopping? (Answer: support $\{0,1,2,\ldots\}$; $P(F_5=0)=0.3277$)
A process requires three successes, with independent success probability $0.25$. Find the probability that it completes by trial 4. (Answer: $P(T_3\le4)=0.0508$)
A process stops at its third success, with independent success probability $0.4$. If it ends on trial 7, state the number of failures before completion and calculate the probability of that exact ending time. (Answer: $F_3=4$; $P(T_3=7)=0.1244$)

Convention	What is counted?	Support	Smallest possible value
\(T_r\)	Trials through and including the \(r\)th success	\(r,r+1,r+2,\ldots\)	\(r\)
\(F_r\)	Failures before the \(r\)th success	\(0,1,2,\ldots\)	\(0\)

What You'll Learn