Geometric as the One-Success Negative Binomial Case

One Success, Two Ways to Count

The geometric distribution describes how long it takes to obtain the first success in a sequence of independent trials with a constant success probability. It is also the negative binomial distribution when the target number of successes is one. The probability model is the same; the main source of confusion is that “waiting time” can count either the trial on which success occurs or only the failures before that trial.

For example, if a hypothetical review process checks applications one at a time and each check independently results in approval with probability $p$, the first approval could occur on the fourth check. That is a total of four checks, but only three failures occurred before approval. Both counts describe the same outcome.

Definition: Suppose trials are independent, each trial has success probability $p$, and $0<p\leq 1$. Write $q=1-p$. Let $T$ be the number of trials up to and including the first success, and let $F$ be the number of failures before the first success. Then $T$ has the trials-count geometric convention, with support $1,2,3,\ldots$, while $F$ has the failures-count convention, with support $0,1,2,\ldots$. They satisfy $T=F+1$.

The support—the set of possible values—is part of a distribution’s definition. A first success cannot occur on trial zero, so $T$ starts at one. But it is possible to have zero failures before that success, so $F$ starts at zero. Their probability mass functions are

P(T=n)=q^{n-1}p,\quad n=1,2,\ldots, \qquad P(F=k)=q^kp,\quad k=0,1,2,\ldots.

These formulas describe the same probabilities with shifted indices: $P(T=n)=P(F=n-1)$. In either case, the event being counted is $n-1$ failures followed by one success, or equivalently $k$ failures followed by one success.

How the Negative Binomial Formula Reduces

The negative binomial model counts trials or failures until a specified target of $r$ successes is reached. Under the trials-count convention, the completion trial $N_r$ has probability function

P(N_r=n)=\binom{n-1}{r-1}p^r q^{n-r}, \quad n=r,r+1,\ldots.

Under the failures-count convention, the number of failures $K_r$ before the $r$th success has probability function

P(K_r=k)=\binom{k+r-1}{r-1}p^r q^k, \quad k=0,1,2,\ldots.

The two counts are related by $N_r=K_r+r$: by the time the $r$th success occurs, exactly $r$ successes and $K_r$ failures have been observed. Setting $r=1$ gives the geometric distribution.

Derivation: Set $r=1$ in the negative binomial trials-count formula. For $n=1,2,\ldots$, there is one way to place the first $r-1=0$ successes among the first $n-1$ trials, so $\binom{n-1}{0}=1$. Therefore $$ P(N_1=n)=\binom{n-1}{0}p q^{n-1}=p q^{n-1}. $$ The failures-count formula gives, for $k=0,1,2,\ldots$, $$ P(K_1=k)=\binom{k}{0}p q^k=pq^k. $$ Thus $N_1=T$, $K_1=F$, and $T=F+1$. The coefficient is one because when the target is one success, there is no choice about the order of earlier successes: every trial before completion must be a failure.

This specialization also gives a direct event-based explanation. For $T=n$, the first $n-1$ trials must all fail and trial $n$ must succeed. Independence makes the probability $q^{n-1}p$. For $F=k$, the first $k$ trials fail and the next succeeds, giving $q^kp$. The formulas are consistent with the support shift, not competing models.

Reading Supports and Events Correctly

A phrase such as “the first success is on the fifth trial” specifies $T=5$, or equivalently $F=4$. A phrase such as “four failures occur before the first success” also specifies $F=4$, so the success is on trial five. In contrast, “there are five failures before the first success” describes $F=5$, and the success is on trial six.

Quantity counted	Symbol	Support	Probability at the displayed value
Trials through the first success	$T$	1, 2, 3, …	$P(T=n)=q^{n-1}p$
Failures before the first success	$F$	0, 1, 2, …	$P(F=k)=q^kp$
Same outcome	$T=F+1$	Shift by one	$P(T=n)=P(F=n-1)$

Cumulative and tail events must be translated with the same care. For an integer $m\geq 1$, “success by trial $m$” means $T\leq m$, or $F\leq m-1$. “No success in the first $m$ trials” means $T>m$, or $F\geq m$. The corresponding probabilities are

P(T\leq m)=P(F\leq m-1)=1-q^m, \qquad P(T>m)=P(F\geq m)=q^m.

The complement in the first expression is that all $m$ trials fail. In the second expression, all $m$ initial trials must fail before the first success can occur later. These expressions use the assumptions of independent trials and the same fixed success probability $p$ on each trial.

Worked Example: Approval on a Specified Trial

In a hypothetical application review process, each application independently receives approval with probability $p=0.25$. Let $T$ be the number of reviews through the first approval, and let $F$ be the number of rejections before that approval. Find (a) the probability that approval occurs on review four, and (b) the probability that approval occurs by review four. State the equivalent failures-count events.

Method. Use the trials-count formula for the exact event and the all-failures complement for the cumulative event. Here $q=1-p=1-0.25=0.75$.

(a) Approval on review four is $T=4$. Equivalently, there are three rejections before approval, so $F=3$. Thus

$$ P(T=4)=P(F=3)=(0.75)^3(0.25) =0.421875(0.25)=0.10546875. $$

To four decimal places, the probability is $0.1055$.

(b) Approval by review four is $T\leq 4$, equivalently $F\leq 3$. The complement is four consecutive rejections:

P(T\leq4)=P(F\leq3)=1-(0.75)^4 =1-0.31640625=0.68359375.

To four decimal places, the probability is $0.6836$. As a check, the exact probabilities of approval on reviews one through four sum to $0.25+0.1875+0.140625+0.10546875=0.68359375$, agreeing with the complement calculation.

Common mistake avoided: The event $T=4$ does not mean four rejections before approval. Four prior rejections would mean $F=4$, and approval would then occur on review five.

Interval Events Under Either Convention

For an interval of possible completion trials, convert both endpoints rather than shifting only one. For example, $a\leq T\leq b$ is equivalent to $a-1\leq F\leq b-1$. One can then add the probabilities for the possible values or subtract cumulative probabilities. In the trials convention, the cumulative formula is $P(T\leq m)=1-q^m$; the same event expressed in failures is $P(F\leq m-1)=1-q^m$.

The one-success case is also a useful check on negative binomial expressions. When $r=1$, every binomial coefficient in the general mass function becomes one. If an alleged one-success formula still contains a nontrivial combinatorial factor, or has a support beginning at one when it counts failures, the convention or substitution should be checked.

Worked Example: A Range of Completion Times

A hypothetical sensor sends independent test signals. Each signal is detected with probability $p=0.60$. Let $T$ be the signal number of the first detection and $F$ the number of undetected signals before it. Find the probability that the first detection is on signal three, four, or five. Express the event in both conventions.

Method. Translate the interval first, then add the negative binomial probabilities specialized to $r=1$. Here $q=1-0.60=0.40$. The trials event $3\leq T\leq5$ is the failures event $2\leq F\leq4$.

\begin{aligned} P(3\leq T\leq5) &=P(2\leq F\leq4)\\ &=P(F=2)+P(F=3)+P(F=4)\\ &=(0.40)^2(0.60)+(0.40)^3(0.60)+(0.40)^4(0.60)\\ &=0.096+0.0384+0.01536\\ &=0.14976. \end{aligned}

To four decimal places, the probability is $0.1498$. As an independent check using cumulative probabilities,

P(T\leq5)-P(T\leq2) =(1-0.40^5)-(1-0.40^2) =0.98976-0.84 =0.14976.

Common mistake avoided: The interval $3\leq T\leq5$ corresponds to $2\leq F\leq4$, not $3\leq F\leq5$. Shifting both endpoints down by one preserves the same outcomes.

Means Also Reflect the Shift

The support shift has a corresponding mean shift. For the geometric distribution with $0<p\leq1$, the mean number of trials through the first success is $E[T]=1/p$, while the mean number of preceding failures is $E[F]=q/p$. Since $T=F+1$, these expressions agree because $q/p+1=(1-p)/p+1=1/p$. They are not two different answers for the same count: one includes the successful trial and the other does not.

This identity is useful when a problem gives an expected waiting count but leaves the wording implicit. Confirm whether the count includes the successful trial before selecting the formula. The formulas require the same independent, constant-$p$ trial model; they do not apply unchanged when success probabilities vary by trial or outcomes are dependent.

Worked Example: Interpreting an Expected Wait

In a hypothetical calibration procedure, independent attempts each succeed with probability $p=0.30$. Let $T$ count attempts through the first success and $F$ count failed attempts before it. Find both expected counts and the probability that success takes at least four attempts. Use the one-success negative binomial interpretation.

Method. Specialize the negative binomial model to one success, then apply the appropriate expectation formula to each support. Here $q=1-0.30=0.70$.

For the number of attempts including the successful attempt,

E[T]=\frac{1}{p}=\frac{1}{0.30}=\frac{10}{3}\approx3.3333.

For the number of failures before success,

E[F]=\frac{q}{p}=\frac{0.70}{0.30}=\frac{7}{3}\approx2.3333.

The difference is exactly one attempt, as required by $T=F+1$. Success taking at least four attempts is $T\geq4$, which means at least three failures before success, $F\geq3$. This requires the first three attempts to fail:

P(T\geq4)=P(F\geq3)=q^3=(0.70)^3=0.3430.

Common mistake avoided: Reporting $7/3$ as the expected number of attempts would omit the successful attempt. It is the expected number of failures before success; the expected total attempts is $10/3$.

Common Convention Errors

Starting a failure count at one. Zero failures is possible, so the failures-count support begins at zero.
Using the same index after changing conventions. $T=n$ corresponds to $F=n-1$, not $F=n$.
Applying the wrong exponent. The exact-event exponents are $n-1$ for $T=n$ and $k$ for $F=k$; in both cases the exponent counts failures.
Confusing “by” with “after.” “By trial $m$” is $T\leq m$, while “no success in the first $m$” is $T>m$. The corresponding failure counts are $F\leq m-1$ and $F\geq m$, respectively.
Forgetting the model assumptions. The geometric distribution requires independent trials with a fixed success probability. A changing probability or dependence between outcomes changes the model.

Key takeaway: The geometric distribution is the negative binomial distribution with target $r=1$. Decide whether the variable counts trials through success or failures before success, state its support, and then use $T=F+1$ to translate the event and its endpoints.

Check Your Understanding

For each problem, identify whether the count includes the successful trial before choosing a formula.

Independent trials succeed with probability $0.20$. If $T$ counts trials through the first success, find $P(T=3)$. (Answer: $0.1280$)
Each independent inspection passes with probability $0.40$. If $F$ counts failures before the first pass, find $P(F\leq2)$. (Answer: $0.7840$)
Independent signals are received successfully with probability $0.30$. Find the probability that the first success occurs on trial two, three, or four. (Answer: $0.4599$)
A negative binomial variable counts failures before the target number of successes. Set the target to one, use success probability $0.25$, and find the probability of exactly four failures. State the corresponding completion trial. (Answer: $0.0791$, with completion on trial 5)
Each independent attempt succeeds with probability $0.50$. Find the probability that there is no success in the first three attempts, expressing the event using both $T$ and $F$. (Answer: $P(T>3)=P(F\geq3)=0.1250$)

Quantity counted	Symbol	Support	Probability at the displayed value
Trials through the first success	\(T\)	1, 2, 3, …	\(P(T=n)=q^{n-1}p\)
Failures before the first success	\(F\)	0, 1, 2, …	\(P(F=k)=q^kp\)
Same outcome	\(T=F+1\)	Shift by one	\(P(T=n)=P(F=n-1)\)

What You'll Learn