The negative binomial distributionEdexcel A-Level Further Maths: Revision notes

The negative binomial distribution extends the geometric distribution. You carry out independent trials, each with the same probability $p$ of success, and let $X$ be the number of trials needed to obtain $r$ successes. We write $X\sim\mathrm{NB}(r,p)$. The conditions are the same as for the geometric model: two outcomes, constant probability $p$ and independent trials. Because the $r$th success must happen on trial $X$, the smallest possible value is $x=r$, so $x=r,r+1,r+2,\dots$ with no upper limit. When $r=1$ it is the geometric distribution. For example, the number of rolls of a fair die needed to get two sixes is $\mathrm{NB}\left(2,\frac16\right)$.

For $X\sim\mathrm{NB}(r,p)$: $\mathrm{P}(X=x)=\binom{x-1}{r-1}p^r(1-p)^{x-r},\quad x=r,r+1,r+2,\dots$ The reasoning gives the formula. Trial $x$ must be the $r$th success, giving a factor $p$. The first $x-1$ trials contain exactly $r-1$ successes in any order, which is $\binom{x-1}{r-1}p^{r-1}(1-p)^{x-r}$. Multiplying gives the formula. Worked example: a shooter hits with probability $0.3$ and $X$ is the number of shots needed for $3$ hits. $\mathrm{P}(X=5)=\binom42(0.3)^3(0.7)^2=6\times0.027\times0.49=0.0794$. For 'on or before' probabilities, add the individual terms: $\mathrm{P}(X\leq4)=\mathrm{P}(X=3)+\mathrm{P}(X=4)=0.027+0.0567=0.0837$. For 'more than', use $1-\mathrm{P}(X\leq x)$ as the number of terms is then smaller.

For $X\sim\mathrm{NB}(r,p)$: $\mu=\mathrm{E}(X)=\frac rp,\qquad \sigma^2=\mathrm{Var}(X)=\frac{r(1-p)}{p^2}.$ These are in the formulae booklet and you do not need to prove them. They are $r$ times the geometric mean and variance, which fits the idea of waiting for $r$ successes one after another. Worked example: $r=4$, $p=0.25$ gives $\mu=\frac{4}{0.25}=16$ and $\sigma^2=\frac{4\times0.75}{0.0625}=48$. For the die with $r=2$ and $p=\frac16$, $\mu=12$ and $\sigma^2=\frac{2\times\frac56}{\frac1{36}}=60$.

State what a trial and a success are, then give $X\sim\mathrm{NB}(r,p)$ with the values of $r$ and $p$. The model suits a 'first to $r$' situation. For a best-of-five series ending when a team has $3$ wins, imagine the games continuing regardless. Then a team wins the series if its third win comes by game $5$, that is $X\leq5$, and you add $\mathrm{P}(X=3)$, $\mathrm{P}(X=4)$ and $\mathrm{P}(X=5)$. For the series to last exactly five games, either team's third win is on game 5: add the two negative binomial probabilities, which cannot happen together. Check an assumption such as independence and constant $p$ when asked to comment on the model.