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How do the geometric and negative binomial distributions model waiting times until successes?

The geometric distribution as a model for the trial of the first success, the negative binomial distribution for the rth success, and their means and variances.

A focused answer to the Edexcel A-Level Further Mathematics Further Statistics content on the geometric and negative binomial distributions, covering the geometric model for the trial of the first success, the negative binomial model for the rth success, and the means and variances of both distributions.

Generated by Claude Opus 4.812 min answer

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  1. What this dot point is asking
  2. The geometric distribution
  3. The negative binomial distribution
  4. Examples in context
  5. Try this

What this dot point is asking

Edexcel Further Statistics wants you to model the number of trials up to and including the first success with the geometric distribution, and up to and including the rrth success with the negative binomial distribution, and to state and use their means and variances. Both are "waiting time" models built on independent trials with a constant success probability.

The geometric distribution

The geometric distribution models how many independent Bernoulli trials it takes to obtain the first success. The variable XX takes values 1,2,3,…1, 2, 3, \ldots, with X=xX = x meaning x−1x - 1 failures followed by a success on the xxth trial. A useful property is the simple tail probability P(X>x)=(1−p)xP(X > x) = (1 - p)^x, which makes "at least" and "more than" questions quick to answer.

The negative binomial distribution

The negative binomial generalises the geometric to the number of trials needed for the rrth success. The variable XX takes values r,r+1,r+2,…r, r + 1, r + 2, \ldots, with X=xX = x meaning the rrth success occurs on the xxth trial. The binomial coefficient (x−1r−1)\binom{x - 1}{r - 1} counts the ways to arrange the first r−1r - 1 successes among the first x−1x - 1 trials, with the last trial fixed as the rrth success.

Examples in context

The geometric and negative binomial distributions extend the discrete-distribution toolkit. They share their foundation (independent trials, constant pp) with the binomial of the poisson-and-binomial dot point, but where the binomial fixes the number of trials and counts successes, these fix the number of successes and count trials, hence "waiting time" models. Their means and variances are derived using the expectation and variance machinery of the discrete-distributions dot point, and the negative binomial coefficient (x−1r−1)\binom{x-1}{r-1} uses the combinatorics from further algebra. Goodness of fit testing (chi-squared) can check whether observed waiting times follow a geometric model.

Try this

Q1. For X∼Geo⁡(0.25)X \sim \operatorname{Geo}(0.25), find the mean. [1 mark]

  • Cue. E(X)=10.25=4E(X) = \frac{1}{0.25} = 4.

Q2. For the same XX, find P(X>2)P(X > 2). [2 marks]

  • Cue. P(X>2)=(0.75)2=0.5625P(X > 2) = (0.75)^2 = 0.5625.

Q3. For a negative binomial with r=3r = 3 and p=0.5p = 0.5, find the variance. [2 marks]

  • Cue. Var⁡(X)=r(1−p)p2=3(0.5)0.25=6\operatorname{Var}(X) = \frac{r(1 - p)}{p^2} = \frac{3(0.5)}{0.25} = 6.

Exam-style practice questions

Practice questions written in the style of Pearson Edexcel exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Edexcel 20196 marksA spinner lands on red with probability 0.30.3 on each spin, independently. Let XX be the number of spins up to and including the first red. Find P(X=4)P(X = 4), E(X)E(X), and P(X≤3)P(X \le 3).
Show worked answer →

Use the geometric pmf, the geometric mean, and the tail probability.

X∼Geo⁡(0.3)X \sim \operatorname{Geo}(0.3), so P(X=4)=(0.7)3(0.3)=0.343×0.3=0.1029P(X = 4) = (0.7)^3(0.3) = 0.343 \times 0.3 = 0.1029 (M1 A1).

E(X)=1p=10.3=3.33E(X) = \frac{1}{p} = \frac{1}{0.3} = 3.33 (3 s.f.) (M1 A1).

P(X≤3)=1−P(X>3)=1−(0.7)3=1−0.343=0.657P(X \le 3) = 1 - P(X > 3) = 1 - (0.7)^3 = 1 - 0.343 = 0.657 (M1 A1).

Edexcel 20226 marksA basketball player scores each free throw independently with probability 0.60.6. Let YY be the number of throws up to and including the third successful throw. Find P(Y=5)P(Y = 5) and state E(Y)E(Y) and Var⁡(Y)\operatorname{Var}(Y).
Show worked answer →

Use the negative binomial pmf with r=3r = 3, then the mean and variance formulae.

YY is negative binomial with r=3r = 3, p=0.6p = 0.6. P(Y=5)=(5−13−1)p3(1−p)5−3=(42)(0.6)3(0.4)2P(Y = 5) = \binom{5 - 1}{3 - 1}p^3(1 - p)^{5 - 3} = \binom{4}{2}(0.6)^3(0.4)^2 (M1 A1).

(42)=6\binom{4}{2} = 6, (0.6)3=0.216(0.6)^3 = 0.216, (0.4)2=0.16(0.4)^2 = 0.16, so P(Y=5)=6×0.216×0.16=0.20736P(Y = 5) = 6 \times 0.216 \times 0.16 = 0.20736 (A1).

E(Y)=rp=30.6=5E(Y) = \frac{r}{p} = \frac{3}{0.6} = 5 (M1 A1). Var⁡(Y)=r(1−p)p2=3(0.4)0.36=1.20.36=3.33\operatorname{Var}(Y) = \frac{r(1 - p)}{p^2} = \frac{3(0.4)}{0.36} = \frac{1.2}{0.36} = 3.33 (A1).

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