What does X\simNB(r,p) count?

The number of independent trials needed to obtain r successes.

State the probability function of X\simNB(r,p).

P(X=x)=\binom{x-1}{r-1}p^r(1-p)^{x-r} for x=r,r+1,r+2,\dots

What is the smallest value of X for NB(r,p)?

x=r, when every trial is a success.

Why is the binomial coefficient \binom{x-1}{r-1} and not \binom{x}{r}?

The last trial is fixed as the rth success, so only the first x-1 trials can be arranged.

What is the key difference between binomial and negative binomial?

Binomial: the number of trials is fixed and you count successes. Negative binomial: the number of successes is fixed and you count trials.

What conditions are needed for a negative binomial model?

Two outcomes per trial, constant probability of success, independent trials, and a fixed number of successes r wanted.

How do you find P(X\leq x) for X\simNB(r,p)?

Add P(X=r)+P(X=r+1)+\dots+P(X=x).

The negative binomial distributionEdexcel A-Level Further Maths: Flashcards

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What does $X\sim\mathrm{NB}(r,p)$ count?

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What does $X\sim\mathrm{NB}(r,p)$ count?: The number of independent trials needed to obtain $r$ successes.
State the probability function of $X\sim\mathrm{NB}(r,p)$ .: $\mathrm{P}(X=x)=\binom{x-1}{r-1}p^r(1-p)^{x-r}$ for $x=r,r+1,r+2,\dots$
What is the smallest value of $X$ for $\mathrm{NB}(r,p)$ ?: $x=r$ , when every trial is a success.
Why is the binomial coefficient $\binom{x-1}{r-1}$ and not $\binom{x}{r}$ ?: The last trial is fixed as the $r$ th success, so only the first $x-1$ trials can be arranged.
What is the link between the geometric and negative binomial distributions?: $\mathrm{NB}(1,p)$ is $\mathrm{Geo}(p)$ .
State $\mathrm{E}(X)$ for $X\sim\mathrm{NB}(r,p)$ .: $\frac rp$
State $\mathrm{Var}(X)$ for $X\sim\mathrm{NB}(r,p)$ .: $\frac{r(1-p)}{p^2}$
What is the key difference between binomial and negative binomial?: Binomial: the number of trials is fixed and you count successes. Negative binomial: the number of successes is fixed and you count trials.
$X\sim\mathrm{NB}(3,0.3)$ . Find $\mathrm{P}(X=5)$ .: $\binom42(0.3)^3(0.7)^2=0.0794$
$X\sim\mathrm{NB}(2,\frac16)$ . Find $\mathrm{E}(X)$ .: $\frac{2}{1/6}=12$
What conditions are needed for a negative binomial model?: Two outcomes per trial, constant probability of success, independent trials, and a fixed number of successes $r$ wanted.
How do you find $\mathrm{P}(X\leq x)$ for $X\sim\mathrm{NB}(r,p)$ ?: Add $\mathrm{P}(X=r)+\mathrm{P}(X=r+1)+\dots+\mathrm{P}(X=x)$ .

Exam questions on The negative binomial distribution

A shooter hits the target with probability $0.3$ on each shot, independently of all other shots. Let $X$ be the number of shots needed to score $3$ hits, so that the third hit occurs on shot $X$ .
Find the probability that the third hit is scored on or before the fourth shot.2 marks
A fair six-sided die is rolled repeatedly. Let $X$ be the number of rolls up to and including the roll on which the second six appears.
Find the probability that more than $3$ rolls are needed to obtain the second six.2 marks
A telesales agent makes a sale on each call with probability $0.25$ , independently of all other calls. Let $X$ be the number of calls made up to and including the call on which the fourth sale is made.
State the distribution of $X$ , and find $\mathrm{E}(X)$ and $\mathrm{Var}(X)$ .3 marks

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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).