The Negative Binomial Distribution

Suppose we perform independent Bernoulli trials, each with success probability $p$, and we keep going until we have observed exactly $r$ successes. Let $X$ denote the number of trials required to obtain the $r$-th success. Then $X$ follows a negative binomial distribution with parameters $r$ and $p$, written $X \sim \text{NB}(r, p)$.

For $X$ to equal $k$, two conditions must hold simultaneously:

1. Trial $k$ is a success (so the $r$-th success lands exactly on trial $k$).
2. Exactly $r-1$ of the first $k-1$ trials are successes.

The number of ways to arrange $r-1$ successes among the first $k-1$ trials is $\binom{k-1}{r-1}$, and each such arrangement has probability $p^{\,r-1}(1-p)^{k-r}$. Multiplying by the probability $p$ that trial $k$ itself is a success, we obtain the PMF:

$$P(X = k) = \binom{k-1}{r-1} p^{\,r} (1-p)^{k-r}, \quad k = r,\, r+1,\, r+2,\, \ldots$$

When $r = 1$, this reduces to the geometric PMF $P(X = k) = p(1-p)^{k-1}$, so the negative binomial generalizes the geometric distribution.

Quick example. With $r = 2$ and $p = \dfrac{1}{2}$, the probability that the $2$nd success occurs on the $4$th trial is

$$P(X = 4) = \binom{3}{1}\left(\tfrac{1}{2}\right)^{2}\!\left(\tfrac{1}{2}\right)^{2} = 3 \cdot \tfrac{1}{16} = \tfrac{3}{16}.$$