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Negative binomial distribution

The number of failures before the rth success in repeated independent trials.

Parameters and support

$r$: number of successes being waited for — $r \in {1, 2, \dots}$
$p$: probability of success on one trial — $0 < p \leq 1$
Support: $x \in {0, 1, 2, \dots}$

The formulas

p(x): $p (x) = (x x + r - 1) p^{r} (1 - p)^{x}$
F(x): $F (x) = I_{p} (r, ⌊ x ⌋ + 1)$
I is the regularized incomplete beta function; no elementary closed form exists.
Mean: $E [X] = \frac{r ( 1 - p )}{p}$
Variance: $Var (X) = \frac{r ( 1 - p )}{p ^{2}}$
MGF: $M (t) = (\frac{p}{1 - ( 1 - p ) e ^{t}})^{r}, t < - ln (1 - p)$

Memory hook. It is r independent geometrics stacked, so everything is r times the geometric's: mean r(1−p)/p, variance r(1−p)/p². Unlike the binomial, its variance is LARGER than its mean.

Where the moments come from

The last trial must be the $r$ th success, so the first $x + r - 1$ trials contain exactly $x$ failures: $(x x + r - 1)$ arrangements.
Each arrangement has probability $p^{r} (1 - p)^{x}$ .
Write $X = \sum_{k = 1}^{r} G_{k}$ with independent geometrics $G_{k}$ .
$E [X] = r \cdot \frac{1 - p}{p}$ and, by independence, $Var (X) = r \cdot \frac{1 - p}{p ^{2}}$ .
The MGF is the geometric MGF raised to the $r$ th power.

Worked example

Each policy application is independently approved with probability 0.6. Find the probability that the third approval occurs on the sixth application.

The sixth application being the third approval means 3 failures precede the 3rd success, so $X = 3$ with $r = 3$ .
$P (X = 3) = (3 3 + 3 - 1) (0.6)^{3} (0.4)^{3} = (3 5) (0.6)^{3} (0.4)^{3}$ .
$(3 5) = 10$ , $(0.6)^{3} = 0.216$ , $(0.4)^{3} = 0.064$ .
$P (X = 3) = 10 \times 0.216 \times 0.064 = 0.13824$ .

Answer: 0.13824

The mean, variance, CDF and moment generating function above are re-derived numerically from this distribution’s own density on every test run — summed over the support for a discrete distribution, integrated by quadrature for a continuous one — and compared with the closed forms printed here. A typo on this page fails the build.

Traps

Using the binomial coefficient C(x + r, x) instead of C(x + r − 1, x) — the last trial is pinned as a success and is not free to move.
Counting trials instead of failures, which shifts the mean by r.

Drill this: the Exam P question bank has original questions on this distribution, and the recall trainer builds its prompts from exactly the formulas above.

Negative binomial distribution

Parameters and support

The formulas

Where the moments come from

Worked example

Traps

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