Maths › Further Statistics 1 › Geometric and negative binomial distributions

Geometric and negative binomial distributions

Stop counting successes in a fixed number of trials and start counting trials until a fixed number of successes. The question turns round, and so does the distribution.

Year FMEDEXCEL 9FM0 FS1 3.1, 3.2, 3.3

Builds on The Poisson distribution and Geometric series.

IN THIS TOPIC

  • Model the trial of a first success with the geometric distribution.
  • Extend to the rth success with the negative binomial distribution.
  • Quote and apply the means and variances of both.

COMMON MISCONCEPTION

Waiting for a success is just the binomial distribution read backwards, so the same formulae apply.

Waiting for the first success

The geometric distribution counts the trial on which the first success arrives. That takes x − 1 failures then a success, so P(X = x) = p(1 − p)x-1. The binomial fixes the trials and counts successes. The geometric fixes the successes and counts trials. The two roles have swapped, and no rearrangement of the binomial formula will get you there. Trials still have to be independent with a constant p, but now there is no upper limit on X, and the probabilities decay geometrically, which is where the name comes from.

P(X=x)=p(1-p)x-1,μ=1p,σ2=1-pp2\text{P}(X = x) = p(1 - p)^{x-1}, μ = \frac{1}{p}, σ^{2} = \frac{1 - p}{p^{2}}IN THE FORMULAE BOOKLET

That whole row is in the booklet's table of standard discrete distributions, so you never need to reconstruct the mean or the variance.

The geometric distribution with p = 0.2: a decaying staircase whose mode is 1 and whose mean is 51234mean 5mode 1: the first trialeach bar × 0.8
FIG. 1The geometric distribution with p = 0.2: each bar is 0.8 times the one before, a decaying staircase with the mean 5 far to the right of the mode at 1.

WORKED EXAMPLE

A first success

A spinner lands on red with probability 0.2. Find the probability the first red is on the third spin, and the mean number of spins needed.

P(X = 3) = 0.2 × 0.8² = 0.128.

μ = 1/p = 5 spins, with variance 0.8/0.04 = 20.

The mode is always 1, however small p is. The single likeliest trial for a first success is the first one, even when the average wait is long.

Waiting for the rth success

The negative binomial generalises it. For the rth success to land on trial x, the first x − 1 trials must hold exactly r − 1 successes, and trial x must succeed:

P(X=x)=x-1Cr-1pr(1-p)x-r\text{P}(X = x) = \, ^{x-1}C_{r-1} \, p^{r}(1 - p)^{x-r}IN THE FORMULAE BOOKLET
μ=rp,σ2=r(1-p)p2μ = \frac{r}{p}, σ^{2} = \frac{r(1 - p)}{p^{2}}IN THE FORMULAE BOOKLET

The negative binomial has its own row in the booklet's table, so look the three up rather than storing them. What the booklet cannot do is tell you which distribution the question is describing. The combination counts where the earlier successes sit. The final trial is pinned, never chosen, and that is the detail candidates most often lose. Setting r = 1 recovers the geometric distribution exactly, means and variances included.

The third success on trial 5: two successes anywhere among the first four trials, and trial 5 pinned as a successSSSSSSSSS6 arrangements × 0.4³ × 0.6² = 0.138pinnedtwo successes, six ways
FIG. 2Waiting for the third success: the first x − 1 trials hold exactly two successes in any arrangement, and trial x is pinned as the third.

WORKED EXAMPLE

The third success

Trials succeed independently with probability 0.4. Find the probability the third success occurs on the fifth trial, and the mean number of trials for three successes.

P(X = 5) = ⁴C₂ × 0.4³ × 0.6² = 6 × 0.064 × 0.36 = 0.138.

μ = r/p = 3/0.4 = 7.5 trials, with variance 3 × 0.6/0.16 = 11.25.

The mean is three times the geometric mean, as it must be. Three independent waits, one after another.

GUIDED PRACTICE

Recognising the model

A quality inspector tests items until finding the second faulty one. Faults occur independently with probability 0.1. Find the probability this happens on the fifth item tested.

Show the working

This is negative binomial with r = 2 and p = 0.1.

P(X = 5) = ⁴C₁ × 0.1² × 0.9³ = 4 × 0.01 × 0.729.

= 0.0292. The combination counts the four places the first fault could have taken among the earlier trials.

ASSESSMENT FOCUS

  • Read what is being counted. Trials until a success is geometric or negative binomial; successes in n trials is binomial.
  • The last trial is always a success in these models, so the combination only arranges the earlier ones.
  • For 'at least x trials', use (1 − p) to the power x − 1. No summation is needed.
  • Quote the means and variances straight from the formulae unless a derivation is asked for.

CHECK YOURSELF

A biased coin shows heads with probability 0.25. Find the probability that the first head appears on the fourth toss, and the mean number of tosses required.

Show a hint

Geometric with p = 0.25.

Show the answer

P(X = 4) = 0.25 × 0.75³ = 0.105. The mean is 1/0.25 = 4 tosses.

Geometric: P(X = x) = p(1 − p)x-1, mean 1/p, variance (1 − p)/p².

Negative binomial for the rth success: the combination arranges the earlier successes, and the mean is r/p.

WORKBOOK

Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.

Question workbook

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Question workbook

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Answer book

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7 questions on this topicAnswer them one at a time and mark yourself against the worked answer.Practise this topic

Or read them with their worked answers on the geometric and negative binomial distributions questions page.

CHECK YOUR PROGRESS

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  • Model the trial of a first success with the geometric distribution.
  • Extend to the rth success with the negative binomial distribution.
  • Quote and apply the means and variances of both.

Open the full revision checklist to see every objective in the course in one place.