Interval Probabilities for Negative Binomial Completion Times

Introduction

A negative binomial completion time answers a practical question: how long must a sequence of independent attempts continue until a specified number of successes has occurred? Often the requested probability is not for one exact completion trial, but for a range: the $r$th success may occur between two deadlines, after a given trial, or no later than a specified attempt. The calculation depends on translating that wording into the correct set of completion trials.

The central procedure is to find the negative binomial probability for each possible completion trial in the requested range and add those probabilities. This tutorial develops that procedure and uses complements when the range has no upper endpoint. The treatment is aligned with the SOA VEE Mathematical Statistics syllabus.

Model, Notation, and Support

Assume a sequence of trials in which each trial has two outcomes, success or failure. The trials are independent, and each has the same success probability $p$, where $0<p<1$. Write $q=1-p$. Let $T_r$ be the trial number on which the $r$th success occurs, for a fixed positive integer $r$.

Definition: Under independent Bernoulli trials with constant success probability $p$, the negative binomial completion time $T_r$ is the trial number of the $r$th success. Its support is the integers $r,r+1,r+2,\ldots$, and for $n\ge r$, $$ P(T_r=n)=\binom{n-1}{r-1}p^r q^{n-r}, \qquad q=1-p. $$

The support matters: $T_r$ cannot be less than $r$, since at least $r$ trials are needed to obtain $r$ successes. A value of $n$ is possible because the first $n-1$ trials must contain exactly $r-1$ successes, and trial $n$ must be a success.

For a range of completion trials, the possible values of $T_r$ describe mutually exclusive events. Therefore, their probabilities add. If the requested range is from trial $a$ through trial $b$, inclusive, only trial numbers that are both in that range and in the support contribute.

P(a\le T_r\le b) = \sum_{n=\max(r,a)}^{b} \binom{n-1}{r-1}p^r q^{n-r}, \qquad \text{when } b\ge \max(r,a).

If $b<\max(r,a)$, the range contains no possible completion trial and its probability is zero. When wording excludes an endpoint, adjust the integer range before summing. For example, “after trial $m$, but by trial $b$” means $m+1\le T_r\le b$, not $m\le T_r\le b$.

Why the Probability Formula Holds

For $T_r=n$, the last trial must be a success. Among the first $n-1$ trials, exactly $r-1$ must be successes. There are $\binom{n-1}{r-1}$ ways to choose which of those trials are successes. Each such sequence has $r$ successes and $n-r$ failures, so its probability is $p^r q^{n-r}$. The possible choices describe disjoint sequences, which gives the formula.

The interval formula follows from the same disjointness: $T_r=n$ and $T_r=k$ cannot both occur when $n\ne k$. Thus, to find the probability that completion occurs at any trial in the requested range, add the exact-completion probabilities for those trials.

Key result: For a finite interval, identify every integer completion trial included by the wording, remove any values below $r$, calculate $P(T_r=n)$ for each remaining value, and add. For a range extending indefinitely, it is often simpler to subtract the probability of the excluded finite range from 1.

A Reliable Procedure for Interval Probabilities

Identify the stopping rule.
Define $T_r$ as the trial on which the $r$th success occurs, and state $r$, $p$, and $q=1-p$.

Translate the wording into integers.
For example, “strictly after $m$” begins at $m+1$, while “at least $m$” begins at $m$.

Check the support.
Ignore completion trials below $r$; they have probability zero.

Sum or complement.
For a finite range, add the exact probabilities. For an open-ended range, consider subtracting the complementary finite range from 1.

Check the result.
Confirm that every included endpoint is present exactly once and that the probability lies between 0 and 1.

Worked Calculations

Worked Example: Completion on Either of Two Scans

A hypothetical monitoring device independently records a designated event on each scan with probability $p=0.5$. Let $T_2$ be the scan on which the second event is recorded. Find the probability that it is recorded on scan 3 or scan 4.

Method. These are two exact completion times, so calculate $P(T_2=3)$ and $P(T_2=4)$, then add. Here $r=2$, $p=0.5$, and $q=0.5$.

On scan 3, the first two scans must contain one success, and scan 3 must be a success:

P(T_2=3) =\binom{2}{1}(0.5)^2(0.5)^1 =2(0.25)(0.5) =0.25.

On scan 4, the first three scans must contain one success, and scan 4 must be a success:

P(T_2=4) =\binom{3}{1}(0.5)^2(0.5)^2 =3(0.25)(0.25) =0.1875.

The two completion events are disjoint, so

P(T_2=3\text{ or }T_2=4) =0.25+0.1875 =0.4375.

Check. The event includes two possible completion scans, and both exact probabilities are between 0 and 1; their sum is $0.4375$. The result is not $0.3750$: that value would omit part of the probability for one of the scans. Common mistake avoided: calculating an interval probability without evaluating and adding every included exact completion time.

Worked Example: After One Trial but No Later Than Another

In a hypothetical data-transfer process, each independent packet check passes with probability $p=0.4$. Let $T_3$ be the check on which the third pass occurs. Find the probability that the third pass occurs after check 4 but by check 7.

Method. “After check 4” excludes check 4, while “by check 7” includes check 7. Thus the event is $5\le T_3\le7$. Here $r=3$, $p=0.4$, and $q=0.6$. All three completion checks are at least $r$, so each is in the support.

For completion on check 5:

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

For completion on check 6:

P(T_3=6) =\binom{5}{2}(0.4)^3(0.6)^3 =10(0.064)(0.216) =0.13824.

For completion on check 7:

P(T_3=7) =\binom{6}{2}(0.4)^3(0.6)^4 =15(0.064)(0.1296) =0.124416.

Adding the disjoint possibilities gives

P(4<T_3\le7) =0.13824+0.13824+0.124416 =0.400896.

Rounded to four decimal places, the probability is $0.4009$. Check. Checks 5, 6, and 7 are all included; check 4 is excluded. Common mistake avoided: interpreting “after 4” as including check 4, or dropping the stated upper endpoint.

Worked Example: Completion on a Specified Trial or Later

A hypothetical manufacturing test independently classifies each unit as satisfactory with probability $p=0.2$. Let $T_3$ be the unit number of the third satisfactory unit. Find the probability that this third unit is unit 6 or later.

Method. The event is $T_3\ge6$. Rather than adding infinitely many probabilities, use the complement: $T_3\ge6$ means the third success has not occurred by unit 5. This is the complement of $T_3\le5$. Here $r=3$, $p=0.2$, and $q=0.8$.

The completion times at or before unit 5 that are possible are 3, 4, and 5. Their probabilities are

P(T_3=3)=\binom{2}{2}(0.2)^3 =1(0.008) =0.008,

P(T_3=4)=\binom{3}{2}(0.2)^3(0.8) =3(0.008)(0.8) =0.0192,

P(T_3=5)=\binom{4}{2}(0.2)^3(0.8)^2 =6(0.008)(0.64) =0.03072.

Therefore,

P(T_3\le5)=0.008+0.0192+0.03072=0.05792,

P(T_3\ge6)=1-0.05792=0.94208.

Rounded to four decimal places, the answer is $0.9421$. Check. The complement covers completion on units 3, 4, or 5, leaving unit 6 onward; the two events partition all possibilities. Common mistake avoided: confusing “unit 6 or later” with “after unit 6,” which would instead begin at unit 7.

Endpoint Language and Common Traps

The probability formula is usually not the difficult part of an interval question. The main risk is translating everyday wording into the correct integer bounds. The following table summarizes common forms.

Wording	Event	Calculation approach
Exactly on trial $n$	$T_r=n$	Use one negative binomial probability.
From trial $a$ through $b$, inclusive	$a\le T_r\le b$	Sum from $\max(a,r)$ through $b$.
After trial $m$, by trial $b$	$m<T_r\le b$	Sum from $\max(m+1,r)$ through $b$.
By trial $b$	$T_r\le b$	Sum possible completion trials through $b$.
Trial $m$ or later	$T_r\ge m$	Use a complement or sum the tail, starting at $\max(m,r)$.

A frequent error is to use the number of failures as the exponent in place of the number of failures before completion. At completion trial $n$, there are $n-r$ failures, so the probability factor is $q^{n-r}$, not $q^n$. Another is to use $\binom{n}{r-1}$ instead of $\binom{n-1}{r-1}$: the final trial is already required to be a success, so the $r-1$ earlier successes must be chosen only from the first $n-1$ trials.

For open-ended events, take particular care with the complement. “The $r$th success occurs at trial $m$ or later” is the complement of “it occurs before trial $m$,” namely $T_r\le m-1$. It is not the complement of $T_r\le m$. Writing the event with inequalities before calculating makes this boundary clear.

Key takeaway: First translate the words into an inequality for $T_r$. Then respect its support, calculate the exact negative binomial probabilities for the included trials, and sum or take a complement. The model requires independent trials with the same success probability; it does not apply as stated if the success probability changes across trials or the outcomes are dependent.

Check Your Understanding

For each problem, define the completion time as the trial of the stated success and identify the included integer trials before calculating.

A hypothetical component passes an independent test with probability $0.3$ on each attempt. Find the probability that the second pass occurs on attempt 4 or 5. (Answer: $0.2558$)
A field instrument independently registers a reading with probability $0.5$ per measurement. Find the probability that its third registered reading occurs from measurement 4 through measurement 6, inclusive. (Answer: $0.5313$)
A hypothetical production cycle yields a target item with probability $0.2$, independently from cycle to cycle. Find the probability that the second target item occurs after cycle 3 but no later than cycle 6. (Answer: $0.2406$)
A test independently identifies a specified feature with probability $0.25$ per trial. Find the probability that the fourth identification occurs from trial 7 through trial 9, inclusive. (Answer: $0.1281$)
A hypothetical inspection independently finds a defect with probability $0.6$ per item. Find the probability that the third defective item is found on item 5 or later. (Answer: $0.5248$)

Wording	Event	Calculation approach
Exactly on trial \(n\)	\(T_r=n\)	Use one negative binomial probability.
From trial \(a\) through \(b\), inclusive	\(a\le T_r\le b\)	Sum from \(\max(a,r)\) through \(b\).
After trial \(m\), by trial \(b\)	\(m<T_r\le b\)	Sum from \(\max(m+1,r)\) through \(b\).
By trial \(b\)	\(T_r\le b\)	Sum possible completion trials through \(b\).
Trial \(m\) or later	\(T_r\ge m\)	Use a complement or sum the tail, starting at \(\max(m,r)\).

What You'll Learn