Learn9FM0Option · Further Statistics 1Geometric and negative binomial distributionsThe geometric distribution

Geometric and negative binomial distributions · Subtopic 1 of 1

The Geometric Distribution

How long until the first success? If each trial succeeds with probability $p$ , the first success comes on trial $r$ with probability $(1-p)^{r-1}p$ : $r-1$ failures, then a success.

Cambridge 9709 Paper 5Edexcel 9FM0 Option≈ 30 minRevision 6 · updated 2026-10-27

Section 1 of 14

Learning objectives

Recognise a geometric situation, $X\sim\text{Geo}(p)$ : the number of trials up to and including the first success.
Use $P(X=r)=q^{r-1}p$ .
Use $P(X>r)=q^r$ and $P(X\le r)=1-q^r$ .
Use $E(X)=\dfrac1p$ .

Pause and think

You throw a die until you get a six. What has to happen for the first six to come on the $4$ th throw? What is its probability?

Section 2 of 14

Core idea

Start hereWaiting for a six

How many rolls until your first six? Could be 1, could be 20. Rolling until the first success, with the same chance each time, gives the geometric distribution. Surprisingly, "first roll" is always the single most likely outcome.

The big idea

$X\sim\operatorname{Geo}(p)$ : trial number of the first success. $P(X=r)=(1-p)^{r-1}p$ ; $P(X>r)=(1-p)^r$ ; $E(X)=\frac1p$ .

1. The formulas

Deriving them

$X=r:\ \underbrace{F\cdots F}_{r-1}S$
$r-1$ failures then a success.
$P(X=r)=(1-p)^{r-1}p$
Independent: multiply.
$P(X>r)=(1-p)^r$
More than $r$ trials needed ⇔ the first $r$ all failed.

Geo(p): the first try is always the most likely

p = 0.10

Each bar is

(1-p)

times the one before, so they shrink steadily. Small

p

gives a long tail: you may wait a long time.

Worked exampleRolling for a six

Find $P(X=4)$ , $P(X>4)$ , $P(X\le4)$ , $E(X)$ .

Goal: Values.
Strategy: $p=\frac16$ .

$(\tfrac56)^3\tfrac16=\tfrac{125}{1296}=0.0965$
Exactly 4th.
$(\tfrac56)^4=0.482$
First four fail.
$1-0.482=0.518$
Complement.
$E(X)=6$
$\frac1p$ .

Answer

0.0965

;

0.482

;

0.518

;

6

✓ Check

P(X\le4)=\frac16(1+\frac56+\frac{25}{36}+\frac{125}{216})=0.518

✓.

Common misconception $P(X\ge r)$

P(X\ge r)=P(X>r-1)=(1-p)^{r-1}

Pause and think

$p=0.2$ : P(first success on 3rd trial or later)?

Summary

$q^{r-1}p$ ; $q^r$ for "more than $r$ "; mean $1/p$ .

Geometric distribution: Trials until the first success.

X ~ Geo(0.3): the trial on which the first success happens

P(X=r)=(0.7)^{r-1}(0.3)

. Each bar is

0.7

times the one before, so

r=1

is always the most likely value. The mean is

\dfrac1p=3.\dot3

HistoryWhy "geometric"? The probabilities

p,\ qp,\ q^2p,\ldots

form a geometric sequence with common ratio

q

, which gives the distribution its name. Adding them all is a geometric series with sum

\dfrac p{1-q}=1

, which confirms they make a valid distribution.

Fun factDice have no memory. If you have thrown

10

non-sixes in a row, the expected number of further throws to get a six is still

6

. The geometric distribution is "memoryless": the past failures do not bring a success any closer. It is the mathematical version of the gambler's fallacy.

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