The Geometric Distribution (Edexcel A Level Further Maths): Revision Note

Exam code: 9FM0

Author

Roger B

Last updated

19 June 2025

Conditions for geometric models

What is the geometric distribution?

The geometric distribution models the number of trials needed to reach the first success
- For example, how many times will you have to roll a dice until it lands on a '6' for the first time
The notation for the geometric distribution is $Geo (p)$
- For a random variable $X$ that has the geometric distribution you can write $X ~ Geo (p)$
- $X$ is the number of trials it takes to reach the first success
  - For example, $X = 8$ means the first success occurred on the 8th trial
- $p$ is the fixed probability of success in any one trial

What are the conditions for using a geometric model?

A geometric model can be used for an experiment that satisfies the following conditions:
- The experiment consists of an indefinite number of successive trials
- The outcome of each trial is independent of the outcomes of all other trials
- There are exactly two possible outcomes for each trial (success and failure)
- The probability of success in any one trial ( $p$ ) is constant
Note that these conditions are very similar to the conditions for the binomial distribution
- But for a binomial distribution the number of trials ( $n$ ) is fixed
  - And you count the number of successes
- While for a negative binomial distribution the experiment continues until the first success is achieved
  - And you count the number of trials it takes to reach that first success

When might the conditions not be satisfied?

If asked to criticise a geometric model, you may be able to question whether the trials are really independent
- For example, someone may be repeating an activity until they achieve a success
  - The trials may not be independent because the person gets better from practising the activity
  - This also means the probability of success, $p$ , is not constant
- In order to proceed using the model, you would have to assume that the trials are independent

Examiner Tips and Tricks

Replace the word "trials" with the context (e.g. "flips of a coin") when commenting on conditions and assumptions

Geometric probabilities

What are the probabilities for the geometric distribution?

If $X ~ Geo (p)$ then $X$ has the probability function:
- $P (X = x) = p (x) = p {(1 - p)}^{x - 1}, x = 1, 2, 3, . . .$
- the random variable $X$ is the number of trials needed to get the first success
- $p$ is the constant probability of success in one trial
- $P (X = x)$ is the probability that the first success will occur on the $x th$ trial

Note that that is the product of
- the probability of first getting $x - 1$ failures, ${(1 - p)}^{x - 1}$ ,
- and the probability of getting a success in the $x th$ trial, $p$
Also note that there is no greatest possible $x$
- It could require any number of trials to reach the first success
- However $P (X = x)$ gets closer and closer to zero as $x$ gets larger
Your calculator may allow you to calculate Geometric probabilities directly
- i.e., without having to use the above formula

What are the properties of the geometric distribution?

Note that $P (X = 1), P (X = 2), P (X = 3), P (X = 4), . . . = p, p (1 - p), p {(1 - p)}^{2}, p {(1 - p)}^{3}, . . .$
- This means that the geometric probabilities form a geometric sequence
  - The first term is $p$
  - The common ratio is $(1 - p)$
- This is where the geometric distribution gets its name!

Assuming that $0 < p < 1$ , then it is also true that $0 < (1 - p) < 1$
- This means that $P (X = 1) > P (X = 2) > P (X = 3) > P (X = 4) > . . .$
  - i.e., the probabilities form a decreasing sequence
  - and $P (X = 1)$ is the largest probability in the sequence
  - Therefore $X = 1$ is the mode of the distribution

The geometric distribution has no 'memory'
- It doesn't matter what has happened previously, or how many 'failures' in a row there have been
- The probability of getting a 'success' in any trial is always $p$
- This means that the number of additional trials needed for the first success is not dependent on the number of trials that have already occurred
  - e.g. if 5 (failed) trials have already occurred, the probability of the first success happening after 7 trials is simply the probability of success happening after 2 trials in the first place, i.e. $p (1 - p)$

What are the cumulative probabilities for the geometric distribution?

If $X ~ Geo (p)$ then $X$ has the cumulative geometric distribution:
- $P (X \leq x) = 1 - {(1 - p)}^{x}, x = 1, 2, 3, . . .$
- the random variable $X$ is the number of trials needed to get the first success
- $p$ is the constant probability of success in one trial
- $P (X \leq x)$ is the probability that the first success will occur on or before the $x th$ trial
Your calculator may allow you to calculate Geometric probabilities directly
- i.e., without having to use the above formula

The formula can be proved as follows
- If the first success occurs on or before the $x th$ trial, that means that the first $x$ trials have not all been failures
  - The probability of getting $x$ failures in a row is ${(1 - p)}^{x}$
  - So the probability of that not happening is $1 - {(1 - p)}^{x}$
Alternatively, it can be proved algebraically
- The geometric probabilities form a geometric sequence with first term $p$ and common ratio $1 - p$
- Putting that into the geometric series formula $S_{n} = \frac{a (1 - r^{n})}{1 - r}$ gives

$P (X \leq x) = \sum_{r = 1}^{x} P (X = r) = \frac{p (1 - {(1 - p)}^{x})}{1 - (1 - p)} = \frac{p (1 - {(1 - p)}^{x})}{p} = 1 - {(1 - p)}^{x}$

Because $\lim_{x \to \infty} {(1 - p)}^{x} = 0$ (assuming $0 < p < 1$ ), it follows that

$\sum_{r = 1}^{\infty} P (X = r) = 1 - 0 = 1$

So the sum of all probabilities is equal to 1
- This is a requirement of any probability distribution

Examiner Tips and Tricks

If you forget the formulae in the exam, you can often still do questions using basic probability concepts and geometric series

Worked Example

Joshua is an inspector in a factory. His job is to randomly sample widgets produced by a particular machine, until he finds a widget that has a defect. If he finds a widget with a defect, then the machine must be stopped until a repair procedure has been completed. Given that the probability of a widget being defective is 0.002, find the probability that:

a) the 10th widget that Joshua inspects is the first one that is defective

b) the 250th widget that Joshua inspects is the first one that is defective

c) the 250th widget that Joshua inspects is the first one that is defective, given that the first 240 were not defective

d) Joshua will inspect 250 or fewer widgets before finding the first one that is defective

e) Joshua will need to inspect more than 250 widgets before finding the first one that is defective.

f) State an assumption you have used in calculating the above probabilities.

Geometric mean & variance

What are the mean and variance of the geometric distribution?

If $X ~ Geo (p)$ , then
- The mean of $X$ is $E (X) = μ = \frac{1}{p}$

The variance of $X$ is $Var (X) = σ^{2} = \frac{1 - p}{p^{2}}$
You need to be able to use these formulae to answer questions about the geometric distribution

Examiner Tips and Tricks

If a question gives you the value of the mean or variance, form an equation in $p$ and solve it

Worked Example

Palamedes is rolling a biased dice for which the probability of the dice landing on a '6' is $p$ . The random variable $X$ represents the number of times he needs to roll the dice until a '6' appears for the first time. Given that the standard deviation of $X$ is $2 \sqrt{3}$ , find:

a) the value of $p$ .

b) the mean of $X$

c) $P (X \leq 3)$ .

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The Geometric Distribution (Edexcel A Level Further Maths): Revision Note

Conditions for geometric models

What is the geometric distribution?

What are the conditions for using a geometric model?

When might the conditions not be satisfied?

Geometric probabilities

What are the probabilities for the geometric distribution?

What are the properties of the geometric distribution?

What are the cumulative probabilities for the geometric distribution?

Geometric mean & variance

What are the mean and variance of the geometric distribution?

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

Discrete Probability Distributions

Discrete Probability Distributions

E(X) & Var(X) of Discrete Random Variables

Linear Transformations of DRVs

Problem Solving with DRVs

Poisson & Binomial Distributions

The Poisson Distribution

Poisson Approximations of Binomials

Geometric & Negative Binomial Distributions

The Geometric Distribution

The Negative Binomial Distribution

Probability Generating Functions

Probability Generating Functions

Probability Generating Functions (PGFs)

E(X) & Var(X) of PGFs

PGFs of Standard Distributions

PGFs of Sums & Transformations

Central Limit Theorem

Central Limit Theorem

Central Limit Theorem

Hypothesis Testing

Poisson & Geometric Hypothesis Testing

Poisson Hypothesis Testing

Geometric Hypothesis Testing

Chi Squared Tests

Goodness of Fit

Chi Squared Tests for Standard Distributions

Chi Squared Tests for Contingency Tables

Quality of Tests

Type I & Type II Errors

Size & Power of Test

Author: Roger B