Probability Formulae (Edexcel GCSE Statistics): Revision Note

Exam code: 1ST0

Written by: Roger B

Reviewed by: Dan Finlay

Updated on 18 April 2024

Addition Law & Mutually Exclusive Events

What are mutually exclusive events?

Two events are mutually exclusive if they cannot both occur at the same time
- e.g. 'get an even number' and 'get an odd number' are mutually exclusive when rolling a dice
- 'get an even number' and 'get a multiple of 3' are not mutually exclusive
  - 6 is both an even number and a multiple of 3
If two events are mutually exclusive you can add their probabilities to find the probability of one or the other happening
- If events A and B are mutually exclusive,
  - then $P (A or B) = P (A) + P (B)$
- This is known as the addition law for two mutually exclusive events
  - But it is often referred to as the 'or' rule

What are exhaustive events?

A set of events is called exhaustive if all possible outcomes are included in the set
- e.g. 'heads' and 'tails' is an exhaustive set for flipping a coin
- 'get an even number' and 'get an odd number' is an exhaustive set for rolling a 6-sided dice
  - but so is 'get a 1', 'get a 2', etc., up to 'get a 6'
For an exhaustive set of mutually exclusive events, the sum of all probabilities is equal to 1
- e.g. $P (heads) + P (tails) = 1$ for flipping a coin
- $P (even) + P (odd) = 1$ for rolling a 6-sided dice
  - or $P (1) + P (2) + P (3) + P (4) + P (5) + P (6) = 1$
This 'sums to 1' property can be used to find unknown probabilities
In particular 'event A happens' and 'event A does not happen' are mutually exclusive and exhaustive
- So $P (A) + P (not A) = 1$
- and $P (not A) = 1 - P (A)$
  - i.e., you can find the probability of A not happening by subtracting the probability of it happening from 1

What is the general addition law?

If two events are not mutually exclusive, then you cannot use the 'or' rule to add probabilities
Instead you can use the general addition law formula
- $P (A or B) = P (A) + P (B) - P (A and B)$
  - i.e. the probability that A or B (or both) occur is equal to
    - the sum of the probabilities for A and B
    - minus the probability that A and B both occur
- This law is always true for any two events A and B
The formula can be used to find any one probability if you know the other three
- Substitute in the values you know, and solve for the one you want to know
- e.g. if $P (A) = 0.2$ , $P (A or B) = 0.7$ and $P (A and B) = 0.1$
  - $0.7 = 0.2 + P (B) - 0.1$
  - $0.7 - 0.2 + 0.1 = P (B)$
  - $P (B) = 0.6$

Examiner Tips and Tricks

Make sure that events are mutually exclusive before using the 'or' rule to add probabilities
And make sure that events are exhaustive and mutually exclusive before using the 'sums to 1' rule
- e.g. in a sport where 'win', 'lose' and 'draw' are all options, $P (lose)$ is not equal to $1 - P (win)$
  - because 'win' and 'lose' are not exhaustive

Worked Example

Emilia is using a spinner with blue, yellow, green, red and purple sectors. The probabilities for the different possibilities are given in a table.

Outcome	Blue	Yellow	Green	Red	Purple
Probability		0.2	0.1		0.4

The spinner has an equal chance of landing on blue or red.

(a) Complete the probability table.

The possibilities listed in the table are mutually exclusive and exhaustive
So all the probabilities should add up to 1

1 - 0.2 - 0.1 - 0.4 = 0.3

So the probability that it lands on blue or red is 0.3
As the probabilities of blue and red are equal you can halve this to get each probability

0.3 ÷ 2 = 0.15

Now complete the table.

Outcome	Blue	Yellow	Green	Red	Purple
Probability	0.15	0.2	0.1	0.15	0.4

(b) Find the probability that the spinner lands on green or purple.

As the spinner can not land on green and purple at the same time they are mutually exclusive
This means you can add their probabilities together.

0.1 + 0.4 = 0.5

P(Green or Purple) = 0.5

The probability of not landing on yellow is equal to 1 minus the probability of landing on yellow

1 - 0.2 = 0.8

P(Not Yellow) = 0.8

Independent Events

What are independent events?

Two events are independent if one event occurring (or not occurring) does not affect the probability of the other event occurring (or not occurring)
- It is not always obvious whether two events are independent or not!
If two events are independent you can multiply their probabilities to find the probability of both events occurring
- If events A and B are independent,
  - then $P (A and B) = P (A) \times P (B)$
- This is known as the multiplication law for independent events
  - But it is often referred to as the 'and' rule
The 'and' rule can be extended to more than two events
- e.g. if events A, B and C are independent,
  - then $P (A and B and C) = P (A) \times P (B) \times P (C)$
The 'and rule can be used to test whether two events are independent
- If $P (A and B) = P (A) \times P (B)$ is true, then A and B are independent
- If $P (A and B) = P (A) \times P (B)$ is not true, then A and B are not independent
Note that if A and B are mutually exclusive, then $P (A and B) = 0$
- So A and B cannot also be independent (unless $P (A) = 0$ or $P (B) = 0$ )
If events A and B are independent, then it is also true that
- $P (A | B) = P (A)$ and $P (B | A) = P (B)$
  - See the spec point on 'Conditional Probability'

Examiner Tips and Tricks

Remember that the 'and' rule can be used in both directions
- If you know events are independent you can use it to calculate probabilities
- You can use it to test whether or not events are independent

Worked Example

(a) The probability that Brendan goes canoeing on a given weekend is 0.3. The probability that he is late for work on a Wednesday is 0.1. Given that those two events are independent, find the probability that Brendan goes canoeing this weekend and is late for work on Wednesday.

Because the events are independent, we can multiply the probabilities

$0.3 \times 0.1 = 0.03$

0.03

(b) There are 52 cards in a normal deck of playing cards, with 13 belonging to each suit (diamonds, hearts, clubs, spades). There are also 4 aces in the deck, with one ace belonging to each suit.

A card is drawn from the deck at random.

Determine whether the events 'draw an ace' and 'draw a spade' are independent.

First we need to work out the individual probabilities

$P (ace) = \frac{4}{52} = \frac{1}{13} P (spade) = \frac{13}{52} = \frac{1}{4}$

The event 'draw an ace and draw a spade' is the same as 'draw the ace of spades'
And there is only one ace of spades in the deck

$P (ace and spade) = \frac{1}{52}$

Now we can multiply those together to see if they satisfy the 'and' rule

$P (ace) \times P (spade) = \frac{1}{13} \times \frac{1}{4} = \frac{1}{52}$

That is equal to $P (ace and spade)$ so the events are independent

$P (ace and spade) = P (ace) \times P (spade)$ , so the two events are independent

Conditional Probability

What are conditional events?

Two events are conditional if one event occurring (or not occurring) affects the probability of the other event occurring (or not occurring)
- If two events are conditional then they are not independent
For example 'it is raining today' and 'I see a person with an umbrella on my way to school' are conditional events
- If it is raining than you are more likely to see a person with an umbrella
- If you don't see any people with umbrellas then it is less likely that it is raining
$P (B | A)$ is the notation for the conditional probability of B given A
- This means the probability that event B occurs given that event A has occurred
  - i.e. this is the probability for B if you know that A has happened
  - It may or may not be the same as $P (B)$
Conditional probabilities are often calculated using Venn diagrams, two-way tables or tree diagrams
- See the 'Probability Diagrams' revision note

What is the formula for conditional probability?

The formula for conditional probability is
- $P (B | A) = \frac{P (A and B)}{P (A)}$
  - i.e. the probability that B occurs given that A has occurred
  - is equal to the probability that A and B both occur
  - divided by the probability that A occurs
It is sometimes useful to rearrange this formula into the following form
- $P (A and B) = P (B | A) \times P (A)$

What about conditional probability and independent events?

If events A and B are independent, then
- $P (A | B) = P (A)$ and $P (B | A) = P (B)$
This makes sense from the definition of independent events
- One independent event occurring (or not) does not affect the probability of the other one occurring (or not)
- The probability of A occurring is always the same whether or not B has occurred
  - so the probability of A given B is just $P (A)$
- The probability of B occurring is always the same whether or not A has occurred
  - so the probability of B given A is just $P (B)$
These equations can be used to test whether or not two events are independent
- If $P (A | B) = P (A)$ and $P (B | A) = P (B)$ are both true, then A and B are independent
- If $P (A | B) = P (A)$ and $P (B | A) = P (B)$ are not both true, then A and B are not independent

Examiner Tips and Tricks

The conditional probability formulae are not on the exam formulae sheet
- so you need to remember them

Worked Example

In a sports club, 15% of all members are people over 40 who play croquet.

Two fifths of the members of the club play croquet.

(a) A member of the club who plays croquet is selected at random. What is the probability that the member is over 40?

Start by writing this information as probabilities

Two fifths of the members play croquet

$P (C) = \frac{2}{5} = 0.4$

15% 'play croquet and are over 40'

$P (C and over 40) = 0.15$

We want to know the probability that a member is over 40 given that they play croquet
Use $P (B | A) = \frac{P (A and B)}{P (A)}$

$P (over 40 | C) = \frac{P (C and over 40)}{P (C)} = \frac{0.15}{0.4} = 0.375$

0.375

The probability that a randomly selected member who is over 40 also plays croquet is $\frac{1}{3}$ .

(b) Explain why 'a member is over 40' and 'a member plays croquet' are not independent events.

Write down the new information as a conditional probability

$P (C | over 40) = \frac{1}{3} = 0.333 . . .$

If two events are independent then $P (A | B) = P (A)$
$P (C | over 40)$ is not equal to $P (C)$ , so the events are not independent

$P (C)$ is not equal to $P (C | over 40)$ . Therefore 'a member is over 40' and 'a member plays croquet' are not independent events

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Previous:Risk & BiasNext:Probability Diagrams

Probability Formulae (Edexcel GCSE Statistics): Revision Note

Addition Law & Mutually Exclusive Events

What are mutually exclusive events?

What are exhaustive events?

What is the general addition law?

Independent Events

What are independent events?

Conditional Probability

What are conditional events?

What is the formula for conditional probability?

What about conditional probability and independent events?

Unlock more, it's free!

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

The Collection of Data

Planning & Types of Data

Planning an Enquiry

Types of Data

Population, Sampling & Collecting Data

Population & Sampling

Collecting Data

Processing, Representing & Analysing Data

Tabulation, Diagrams & Representation

Bar Charts, Line Graphs & Pictograms

Pie Charts

Stem & Leaf Diagrams

Two-way Tables & Venn Diagrams

Population Pyramids

Choropleth Maps

Cumulative Frequency Charts

Box Plots

Histograms & Frequency Polygons

Selecting & Interpreting Data Representations

Measures of Central Tendency

Mode, Median & Arithmetic Mean

Mode & Mean from Grouped Data

Linear Interpolation

Transforming Data

Other Types of Mean

Measures of Dispersion

Quartiles & Percentiles

Types of Range

Standard Deviation

Outliers

Using Summary Statistics

Using Measures of Central Tendency

Using Measures of Dispersion

Skewness

Index Numbers & Rates of Change

Index Numbers

Rates of Change

Scatter Diagrams & Correlation

Scatter Diagrams & Correlation

Lines of Best Fit & Regression Lines

Spearman's Rank Correlation Coefficient

Pearson's Product Moment Correlation Coefficient (PMCC)

Time Series

Time Series Graphs

Moving Averages

Identifying & Interpreting Trends in Data

Quality Assurance & Estimation

Quality Assurance

Estimation from Statistical Data

Probability

Probability Basics

Probabilities & Data

Risk & Bias

Probability Formulae

Probability Diagrams

Probability Distributions

The Binomial Distribution

The Normal Distribution

Author: Roger B

Reviewer: Dan Finlay