The geometric distribution counts the trials up to the first success. Waiting for the third success, or the fifth, gives the negative binomial distribution. As in that lesson, $X$ counts all the trials, including the last success. (Some books count only the failures, $X - r$; the ideas are the same.)
Key idea
In independent trials with success probability $p$, the number of trials $X$ needed for the $r$th success has $$P(X = k) = \binom{k - 1}{r - 1}p^{r}(1 - p)^{k - r}, \quad k = r, r + 1, \ldots,$$ with $E[X] = \frac{r}{p}$ and $\operatorname{Var}(X) = \frac{r(1 - p)}{p^{2}}$.
Why it works
$X = k$ means trial $k$ is a success and the first $k - 1$ trials contain exactly $r - 1$ successes, in any of $\binom{k - 1}{r - 1}$ arrangements. Each arrangement has $r$ successes and $k - r$ failures, probability $p^{r}(1 - p)^{k - r}$. For the mean, split the wait into $r$ stages: trials to the first success, then from there to the second, and so on. Each stage is geometric with mean $\frac{1}{p}$, so by linearity $E[X] = \frac{r}{p}$. The stages are independent, so their variances $\frac{1 - p}{p^{2}}$ add too.
Worked examples
Worked example: Coins
Flip a fair coin until the 3rd head. Find the probability that exactly 5 flips are needed.
- The 5th flip is the 3rd head, and the first 4 flips hold exactly 2 heads: $\binom{4}{2} = 6$ arrangements.
- $P(X = 5) = 6 \cdot \left(\frac{1}{2}\right)^{5} = \frac{3}{16}$.
Answer: $\frac{3}{16}$
Worked example: Sixes
Roll a die until the 2nd six. Find $P(X = 4)$ and $E[X]$.
- $$P(X = 4) = \binom{3}{1}\left(\frac{1}{6}\right)^{2}\left(\frac{5}{6}\right)^{2} = 3 \cdot \frac{25}{1296} = \frac{25}{432} \approx 0.058$$
- $E[X] = \frac{2}{1/6} = 12$ rolls.
Answer: $\frac{25}{432}$ and 12
Worked example: A best-of-seven series
Team A wins each game with probability $0.6$, independently. In a best-of-seven series (first to 4 wins), find the probability that A wins in exactly 6 games.
- Game 6 must be A's 4th win, and games 1 to 5 hold exactly 3 A-wins: $\binom{5}{3} = 10$ arrangements.
- $$P = 10(0.6)^{4}(0.4)^{2} = 10(0.1296)(0.16) \approx 0.207$$
Answer: about $0.207$
"The $r$th success comes on or before trial $m$" is the same event as "at least $r$ successes in the first $m$ trials". So you can add the negative binomial probabilities for $k = r, \ldots, m$, or add binomial probabilities for $r$ or more successes in $m$ trials. Both give the same number, a useful check.
Common mistake
Using $\binom{k}{r}$ instead of $\binom{k - 1}{r - 1}$. The last trial is not free: it must be the $r$th success, so only the earlier $k - 1$ trials can be arranged. Check with $r = 1$: the formula must give the geometric $p(1 - p)^{k - 1}$, and $\binom{k - 1}{0} = 1$ does.