The Geometric Distribution As A Negative Binomial Case

Introduction

The geometric distribution models how many failures occur before the first success in a sequence of repeated trials. In a hypothetical claim-screening process, for example, a “success” might mean that a submitted claim meets a specified criterion; the random variable could count how many claims fail that criterion before the first one meets it. The model is appropriate only when the trials are independent and each has the same probability of success.

The geometric distribution is not a separate case unrelated to the negative binomial. Under the failures-count convention, it is precisely the negative binomial distribution with one required success. This connection gives a direct route to its probability formula and helps prevent a common error: counting the successful trial when the variable is supposed to count only failures.

Definition: Suppose independent trials continue until the first success, and each trial has success probability $p$, where $0<p\leq 1$. Let $X$ be the number of failures before that first success. Then $X$ has a geometric distribution with support $x=0,1,2,\ldots$. Write $q=1-p$ for the probability of failure on any one trial.

Geometric as the One-Success Negative Binomial Case

In the negative binomial failures-count model, the random variable counts failures before the $r$th success. To have exactly $x$ failures before the $r$th success, there must be $x+r-1$ trials before the final trial, and the final trial must be a success. The $x$ failures and $r-1$ earlier successes can be arranged among those first $x+r-1$ trials in $\binom{x+r-1}{x}$ ways. Each such sequence has probability $q^x p^r$. Therefore,

P(X=x)=\binom{x+r-1}{x}p^r q^x,\qquad x=0,1,2,\ldots

Set $r=1$. The binomial coefficient becomes $\binom{x}{x}=1$, and $p^r=p$. Thus the geometric probability formula is

P(X=x)=p q^x,\qquad x=0,1,2,\ldots

The factor $q^x$ accounts for the first $x$ failures, and the factor $p$ accounts for the success on the next trial. In particular, $P(X=0)=p$: zero failures means the very first trial is a success. The probabilities sum to one when $0<p\leq1$, since for $0<p<1$,

\sum_{x=0}^{\infty}p q^x=\frac{p}{1-q}=1, \qquad\text{because }1-q=p.

When $p=1$, success occurs on the first trial with certainty, so $X=0$ with probability one. The same point-probability formula gives this degenerate case.

Key distinction: Here $X$ counts failures only, so its support begins at zero. If instead $N$ counts the total number of trials through and including the first success, then $N=X+1$, and its support begins at one. Do not use a trials-count formula when the question asks for failures.

Mean and Variance

The negative binomial mean under this convention is $r q/p$, so setting $r=1$ gives $E[X]=q/p$. The variance can also be obtained from the geometric probability formula. The probability generating function provides a compact derivation of both quantities and makes clear that they follow from the model’s probabilities.

Geometric mean and variance: For the number $X$ of failures before the first success, with success probability $p$ and failure probability $q=1-p$, $$ E[X]=\frac{q}{p},\qquad \operatorname{Var}(X)=\frac{q}{p^2}. $$ These formulas use the failures-count convention and assume independent trials with constant success probability.

To justify the formulas, define the probability generating function by $G_X(z)=E[z^X]$. For $0<p<1$, substituting the geometric probabilities and summing a geometric series gives

G_X(z)=\sum_{x=0}^{\infty}p q^x z^x =\frac{p}{1-qz}.

For a probability generating function, $G_X'(1)=E[X]$ and $G_X''(1)=E[X(X-1)]$. Differentiating and evaluating at $z=1$ gives

G_X'(z)=\frac{pq}{(1-qz)^2}, \qquad G_X''(z)=\frac{2pq^2}{(1-qz)^3}, $$ $$ G_X'(1)=\frac{q}{p}, \qquad G_X''(1)=\frac{2q^2}{p^2}.

Since $X^2=X(X-1)+X$, it follows that

E[X^2]=G_X''(1)+G_X'(1) =\frac{2q^2}{p^2}+\frac{q}{p}.

Subtracting the square of the mean yields the variance:

\operatorname{Var}(X) =E[X^2]-\{E[X]\}^2 =\frac{2q^2}{p^2}+\frac{q}{p}-\frac{q^2}{p^2} =\frac{q^2}{p^2}+\frac{q}{p} =\frac{q(q+p)}{p^2} =\frac{q}{p^2}.

The formulas also hold when $p=1$: then $q=0$, and both the mean and variance are zero, as required for $X=0$ with certainty.

Calculating Probabilities and Moments

For an exact value, substitute into $P(X=x)=p q^x$. For a range, add the relevant point probabilities or use the complement. For example, the event $X\geq m$ means the first $m$ trials are all failures, so $P(X\geq m)=q^m$. This shortcut follows directly from the event description; it does not include a success on trial $m$.

Worked Example: Exact and Cumulative Probabilities

A hypothetical sequence of independent inspections has probability $p=0.40$ that an inspection identifies a specified defect. Let $X$ be the number of inspections that do not identify the defect before the first inspection that does. Find $P(X=2)$, $P(X\geq3)$, the mean, and the variance.

Method. This is a failures-count geometric model with constant success probability $p=0.40$. The failure probability is $q=1-p=0.60$.

For exactly two failures, the first two inspections fail and the third succeeds:

$$ P(X=2)=p q^2 =(0.40)(0.60)^2 =(0.40)(0.36) =0.1440. $$

For at least three failures, the first three inspections must all fail; what happens afterward does not affect this event:

P(X\geq3)=q^3=(0.60)^3=0.2160.

The mean number of failures is

E[X]=\frac{q}{p}=\frac{0.60}{0.40}=1.5.

The variance is

\operatorname{Var}(X)=\frac{q}{p^2} =\frac{0.60}{(0.40)^2} =\frac{0.60}{0.16}=3.75.

Check. The exact probability is less than one, and $P(X\geq3)=0.2160$ agrees with the requirement that three consecutive failures occur. Probabilities are reported to four decimal places; the mean and variance are exact. A common mistake would be to use $p q^3$ for $P(X\geq3)$, which incorrectly requires a success immediately after three failures. The event only requires the first three failures, so its probability is $q^3$.

Worked Example: Recovering the Parameter From the Mean

A hypothetical review process continues until the first item passes a specified test. The number $X$ of failed items before that pass has a geometric distribution, and its mean is known to be $3$. Find the success probability, the variance, and the probability of exactly four failures.

Method. Use the geometric mean to solve for $p$, then use the same value of $p$ in the variance and point-probability formulas. Since $E[X]=q/p$ and $q=1-p$,

3=\frac{1-p}{p}, \qquad 3p=1-p, \qquad 4p=1, \qquad p=0.25.

Thus $q=1-0.25=0.75$. The variance is

\operatorname{Var}(X)=\frac{q}{p^2} =\frac{0.75}{(0.25)^2} =\frac{0.75}{0.0625}=12.

For exactly four failures, there must be four failures followed by a success:

$$ P(X=4)=p q^4 =(0.25)(0.75)^4 =(0.25)(0.31640625) =0.0791. $$

Check. The success probability $0.25$ gives a mean of $0.75/0.25=3$, matching the stated mean. The variance is $12$, and the probability is reported to four decimal places. A common mistake is to treat the mean of three failures as a mean of three trials and set $p=1/3$. The failures-count mean is $q/p$, not $1/p$; the latter is the mean of the total number of trials $X+1$.

Worked Example: Applying the One-Success Negative Binomial Case

A hypothetical monitoring sequence is modeled by the negative binomial failures-count distribution with $r=1$ and success probability $p=0.25$. Find the probability of at most two failures before the first success, and verify that the one-success case gives the same mean and variance as the geometric formulas.

Method. Because $r=1$, this negative binomial variable is geometric. Set $q=1-p=0.75$. The event $X\leq2$ consists of $X=0$, $X=1$, and $X=2$:

P(X\leq2) =p+p q+p q^2 =(0.25)+(0.25)(0.75)+(0.25)(0.75)^2 $$ $$ =0.25+0.1875+0.140625 =0.578125 \approx0.5781.

The negative binomial mean at $r=1$ is

E[X]=\frac{r q}{p} =\frac{(1)(0.75)}{0.25}=3.

Its variance at $r=1$ is

\operatorname{Var}(X)=\frac{r q}{p^2} =\frac{(1)(0.75)}{(0.25)^2}=12.

These match the geometric formulas $q/p=3$ and $q/p^2=12$. As a second check on the cumulative probability, its complement is $P(X\geq3)=q^3=(0.75)^3=0.421875$, and $1-0.421875=0.578125$. The probability is reported to four decimal places. A common mistake is to substitute $r=1$ into the negative binomial formula but retain an unnecessary combination factor other than one: $\binom{x}{x}=1$, so the point probability is simply $p q^x$.

Model Checks and Common Traps

The geometric model is appropriate only if the process can be represented as successive independent Bernoulli trials with the same success probability $p$, continuing until the first success. The formula does not apply just because a variable is a nonnegative count. For example, if the success probability changes from trial to trial, or if the outcome of one trial affects the next, then the geometric probabilities need not hold.

Before calculating, identify what the problem calls a success and what the random variable counts. Then verify that the trials are independent, that $p$ is constant, and that the process stops at the first success. These checks settle the support and prevent an off-by-one error.

Counting the success as a failure. Under this convention, $X=0$ is possible and means success on the first trial.
Using the wrong event probability. Exactly $x$ failures has probability $p q^x$; at least $x$ failures has probability $q^x$.
Mixing the two conventions. If a question counts trials through the success, the count is $X+1$, not $X$.
Using a success count other than one. For $r>1$, the negative binomial combination factor and powers change; the geometric formula is the special case $r=1$.

Key takeaway: With independent trials, constant success probability $p$, and $X$ counting failures before the first success, $X$ is negative binomial with $r=1$: $P(X=x)=p(1-p)^x$, $E[X]=(1-p)/p$, and $\operatorname{Var}(X)=(1-p)/p^2$.

Check Your Understanding

For each problem, use the failures-count convention unless the question explicitly asks for the total number of trials.

A hypothetical inspection succeeds independently with probability $0.20$ on each attempt. Let $X$ count failures before the first success. Find $P(X=0)$, $E[X]$, and $\operatorname{Var}(X)$. (Answer: $0.2000,\ 4,\ 20$)
A screening trial has success probability $0.50$. Find the probability of exactly three failures before success, the probability of at most one failure, and the mean number of failures. (Answer: $0.0625,\ 0.7500,\ 1$)
In a hypothetical repeated test, each trial succeeds with probability $0.30$. Find $P(X=2)$, $P(X\geq3)$, and $\operatorname{Var}(X)$. (Answer: $0.1470,\ 0.3430,\ 7.7778$)
A process has a geometric failures-count mean of $4$. Find its success probability, variance, and probability of exactly one failure. (Answer: $p=0.2000,\ \operatorname{Var}(X)=20,\ P(X=1)=0.1600$)
Independent trials have success probability $0.40$. Find the probability of at most two failures before the first success, and then give the mean number of total trials through and including success. (Answer: $0.7840,\ 2.5$ trials)

What You'll Learn