Probability

What you'll learn

How to describe events using probability notation and Venn diagrams.
How to recognise mutually exclusive and independent events.
How to use tree diagrams for multi-stage problems.
How to combine “and” and “or” probabilities accurately.

Probability basics

Probability measures how likely something is to happen. It is always between 0 and 1, where 0 means impossible and 1 means certain.

Definition

Event and sample space

The sample space is the set of all possible outcomes.
An event is a particular outcome, or group of outcomes, from the sample space.
The complement of event $A$ , written $A^{'}$ , means “not $A$ ”.

If all outcomes are equally likely, then:

outcomesP(A)=\frac{\text{number of favourable outcomes}}{\text{total number of equally likely outcomes}}

Also remember:

\le P(A) \le 1

and

P (A^{'}) = 1 - P (A)

Example

Using a complement

A bag contains 7 blue counters, 5 red counters and 3 green counters. One counter is chosen at random. Find the probability that it is not red.

A bag model showing that the complement of red includes the blue and green counters.

Find the total number of counters:
$7 + 5 + 3 = 15$
Find the probability of choosing a red counter:
$P(R)=515=13P(R)=\frac{5}{15}=\frac{1}{3}$
Use the complement rule for “not red”:
$P(R′)=1−13=23P(R')=1-\frac{1}{3}=\frac{2}{3}$

Venn diagrams and set notation

A Venn diagram shows events as circles inside a rectangle. The rectangle represents the universal set: everything being considered.

The overlap between circles represents outcomes that belong to more than one event.

Venn diagram showing the regions formed by three overlapping events A, B and C

Definition

Union and intersection

$\cap B$ means A and B: the overlap of the two events.
$\cup B$ means A or B: everything in A, in B, or in both.
“Or” in probability is inclusive, so it includes the overlap unless told otherwise.

Key Idea

The two-event Venn formula

For any two events:

\cup B)=P(A)+P(B)-P(A \cap B)

You subtract the overlap because it has been counted twice.

Example

Filling a two-event Venn diagram

Events $A$ and $B$ have $P (A) = 0.6$ , $P (B) = 0.5$ , and the probability of neither event is 0.1. Find $\cap B)$ and the probability in each region.

A completed two-event Venn diagram with the four probabilities placed in their regions.

First find the probability of $\cup B$ , since “neither” is outside both circles:
$\cup B)=1-0.1=0.9$
Use the Venn formula to find the overlap:
$\cap B)=0.6+0.5-0.9=0.2$
Find the part in $A$ only:
$0.6 - 0.2 = 0.4$
Find the part in $B$ only:
$0.5 - 0.2 = 0.3$
Check that all regions add to 1:
$0.4 + 0.2 + 0.3 + 0.1 = 1$

Common Mistake

Double-counting the overlap

When finding $\cup B)$ , do not just add $P (A)$ and $P (B)$ unless the events have no overlap. If they overlap, the shared region has been counted twice.

Mutually exclusive events

Definition

Mutually exclusive events

Two events are mutually exclusive if they cannot happen at the same time.

For mutually exclusive events:

\cap B)=0

For example, if one counter is chosen from a bag, the events “it is red” and “it is blue” are mutually exclusive.

Example

Using mutually exclusive events

In a Venn diagram, event $C$ does not overlap event $A$ . Given $P (A) = 0.45$ and $P (C) = 0.25$ , find $\cup C)$ .

Disjoint circles show that mutually exclusive events have no overlap.

Since $A$ and $C$ do not overlap, their intersection is zero:
$\cap C)=0$
Use the union formula:
$\cup C)=P(A)+P(C)-P(A \cap C)$
Substitute the values:
$\cup C)=0.45+0.25-0=0.70$

Common Mistake

Mutually exclusive is not the same as independent

If two events are mutually exclusive and both have positive probability, then knowing one has happened makes the other impossible. That means they are not independent.

Three-set Venn diagrams

For three events, fill the diagram from the middle outwards.

The most important point is that pairwise totals such as “football and golf” usually include people who also chose the third option.

Tip

Middle first

In a three-set Venn diagram, put the “all three” value in the centre first. Then subtract it from each two-way overlap before finding the “only” regions.

Example

Completing a three-set Venn diagram

A group of 50 people are asked which sports they watch: football, golf and hockey.

A completed three-set Venn diagram showing the centre-first method for football, golf and hockey.

6 watch all three sports.
10 watch football and golf.
9 watch golf and hockey.
8 watch football and hockey.
28 watch football, 18 watch golf and 20 watch hockey.

Complete the Venn diagram and find the probability that two people chosen at random both watch football.

Put the “all three” value in the centre:
$6$
Subtract the centre from each pairwise total:
$only=8−6=2\begin{aligned} \text{football and golf only} &= 10-6=4\\ \text{golf and hockey only} &= 9-6=3\\ \text{football and hockey only} &= 8-6=2 \end{aligned}$
Find each “only” region:
$only=20−2−3−6=9\begin{aligned} \text{football only} &= 28-4-2-6=16\\ \text{golf only} &= 18-4-3-6=5\\ \text{hockey only} &= 20-2-3-6=9 \end{aligned}$
Find how many people watch none of the three sports:
$50 - (16 + 5 + 9 + 4 + 3 + 2 + 6) = 5$
For two people both watching football, sample without replacement:
$2850×2749=3781225≈0.309\frac{28}{50}\times\frac{27}{49}=\frac{378}{1225}\approx 0.309$

Independent events

Definition

Independent events

Two events are independent if the outcome of one event does not affect the probability of the other.

For independent events:

\cap B)=P(A)P(B)

This formula is the test you should use when asked whether two events are statistically independent.

Example

Testing for independence

A completed Venn diagram has these regions: $A$ only is 0.28, $B$ only is 0.18, the overlap is 0.12, and outside both is 0.42. Decide whether $A$ and $B$ are independent.

The Venn diagram for testing independence by comparing the overlap with the product of the event probabilities.

Find $P (A)$ by adding the regions inside circle $A$ :
$P (A) = 0.28 + 0.12 = 0.40$
Find $P (B)$ by adding the regions inside circle $B$ :
$P (B) = 0.18 + 0.12 = 0.30$
Compare the actual overlap with the product:
$P(A)P(B)=0.40×0.30=0.12P(A)P(B)=0.40\times 0.30=0.12$
Since this equals the overlap, the events are independent:
$\cap B)=P(A)P(B)$

Tip

How to write the reason

For independence questions, write a full reason such as: “Yes, because $\cap B)=P(A)P(B)$ .” A bare “yes” is not enough.

Tree diagrams

A tree diagram shows a probability experiment in stages. Each branch has a probability.

For staged events:

Multiply along branches for “and”.
Add separate routes for “or”.
If there is no replacement, update the numbers after the first choice.

Probability tree diagram for two counters drawn without replacement

Definition

Conditional probability

A conditional probability is the probability of an event given that something else has already happened.

It is written as $P(B∣A)P(B\mid A)$ , meaning “the probability of $B$ given $A$ ”.

Example

Two items without replacement

A box contains 10 milk chocolates and 8 dark chocolates. Connor takes two chocolates at random without replacement. Find the probability that he takes two dark chocolates, and the probability that he takes one of each type.

A two-stage tree diagram shows how probabilities change after the first chocolate is taken without replacement.

For two dark chocolates, multiply down the dark-then-dark route:
$P(DD)=818×717=28153P(DD)=\frac{8}{18}\times\frac{7}{17}=\frac{28}{153}$
For one of each type, there are two possible orders: milk then dark, or dark then milk.
$P(MD)=1018×817=40153P(DM)=818×1017=40153\begin{aligned} P(MD)&=\frac{10}{18}\times\frac{8}{17}=\frac{40}{153}\\ P(DM)&=\frac{8}{18}\times\frac{10}{17}=\frac{40}{153} \end{aligned}$
Add the two routes:
$each)=40153+40153=80153P(\text{one of each})=\frac{40}{153}+\frac{40}{153}=\frac{80}{153}$

Common Mistake

Forgetting the second order

“One of each” usually has two routes: first type A then type B, or first type B then type A. Work out both unless the question fixes the order.

Total probability

Sometimes an outcome can happen through several different routes. Find each route probability, then add them.

Example

Faulty components from different machines

A factory uses three machines. Machine A makes 40% of the components, machine B makes 35%, and machine C makes 25%. The faulty rates are 3% for A, 2% for B and 1% for C. Find the probability that a randomly selected component is from A and faulty, then find the probability that it is faulty.

A total probability tree shows the separate routes through machines A, B and C to a faulty component.

Convert the percentages to decimals before multiplying.
For “from A and faulty”, multiply along that route:
$faulty)=0.40×0.03=0.012P(\text{A and faulty})=0.40\times 0.03=0.012$
A faulty component can come from A, B or C, so add the three routes:
$P(faulty)=(0.40×0.03)+(0.35×0.02)+(0.25×0.01)P(\text{faulty})=(0.40\times 0.03)+(0.35\times 0.02)+(0.25\times 0.01)$
Calculate the final probability:
$P(faulty)=0.012+0.007+0.0025=0.0215P(\text{faulty})=0.012+0.007+0.0025=0.0215$

Exam technique

In the exam

Label every region or branch clearly before doing calculations.
For Venn diagrams, remember: all regions must add to 1 for probabilities, or to the total number of people/items for frequencies.
For tree diagrams, multiply along a route and add different routes; check whether the experiment is with or without replacement.

Self review

Check yourself

Can you explain the difference between mutually exclusive and independent events?
In a three-set Venn diagram, why do you fill the centre first?
When a question asks for “one of each”, have you considered both possible orders?

Probability

What you'll learn

How to describe events using probability notation and Venn diagrams.
How to recognise mutually exclusive and independent events.
How to use tree diagrams for multi-stage problems.
How to combine “and” and “or” probabilities accurately.

Probability basics

Probability measures how likely something is to happen. It is always between 0 and 1, where 0 means impossible and 1 means certain.

Definition

Event and sample space

The sample space is the set of all possible outcomes.
An event is a particular outcome, or group of outcomes, from the sample space.
The complement of event $A$ , written $A^{'}$ , means “not $A$ ”.

If all outcomes are equally likely, then:

outcomesP(A)=\frac{\text{number of favourable outcomes}}{\text{total number of equally likely outcomes}}

Also remember:

\le P(A) \le 1

and

P (A^{'}) = 1 - P (A)

Example

Using a complement

A bag contains 7 blue counters, 5 red counters and 3 green counters. One counter is chosen at random. Find the probability that it is not red.

A bag model showing that the complement of red includes the blue and green counters.

Find the total number of counters:
$7 + 5 + 3 = 15$
Find the probability of choosing a red counter:
$P(R)=515=13P(R)=\frac{5}{15}=\frac{1}{3}$
Use the complement rule for “not red”:
$P(R′)=1−13=23P(R')=1-\frac{1}{3}=\frac{2}{3}$

Venn diagrams and set notation

A Venn diagram shows events as circles inside a rectangle. The rectangle represents the universal set: everything being considered.

The overlap between circles represents outcomes that belong to more than one event.

Venn diagram showing the regions formed by three overlapping events A, B and C

Definition

Union and intersection

$\cap B$ means A and B: the overlap of the two events.
$\cup B$ means A or B: everything in A, in B, or in both.
“Or” in probability is inclusive, so it includes the overlap unless told otherwise.

Key Idea

The two-event Venn formula

For any two events:

\cup B)=P(A)+P(B)-P(A \cap B)

You subtract the overlap because it has been counted twice.

Example

Filling a two-event Venn diagram

Events $A$ and $B$ have $P (A) = 0.6$ , $P (B) = 0.5$ , and the probability of neither event is 0.1. Find $\cap B)$ and the probability in each region.

A completed two-event Venn diagram with the four probabilities placed in their regions.

First find the probability of $\cup B$ , since “neither” is outside both circles:
$\cup B)=1-0.1=0.9$
Use the Venn formula to find the overlap:
$\cap B)=0.6+0.5-0.9=0.2$
Find the part in $A$ only:
$0.6 - 0.2 = 0.4$
Find the part in $B$ only:
$0.5 - 0.2 = 0.3$
Check that all regions add to 1:
$0.4 + 0.2 + 0.3 + 0.1 = 1$

Common Mistake

Double-counting the overlap

When finding $\cup B)$ , do not just add $P (A)$ and $P (B)$ unless the events have no overlap. If they overlap, the shared region has been counted twice.

Mutually exclusive events

Definition

Mutually exclusive events

Two events are mutually exclusive if they cannot happen at the same time.

For mutually exclusive events:

\cap B)=0

For example, if one counter is chosen from a bag, the events “it is red” and “it is blue” are mutually exclusive.

Example

Using mutually exclusive events

In a Venn diagram, event $C$ does not overlap event $A$ . Given $P (A) = 0.45$ and $P (C) = 0.25$ , find $\cup C)$ .

Disjoint circles show that mutually exclusive events have no overlap.

Since $A$ and $C$ do not overlap, their intersection is zero:
$\cap C)=0$
Use the union formula:
$\cup C)=P(A)+P(C)-P(A \cap C)$
Substitute the values:
$\cup C)=0.45+0.25-0=0.70$

Common Mistake

Mutually exclusive is not the same as independent

If two events are mutually exclusive and both have positive probability, then knowing one has happened makes the other impossible. That means they are not independent.

Three-set Venn diagrams

For three events, fill the diagram from the middle outwards.

The most important point is that pairwise totals such as “football and golf” usually include people who also chose the third option.

Tip

Middle first

In a three-set Venn diagram, put the “all three” value in the centre first. Then subtract it from each two-way overlap before finding the “only” regions.

Example

Completing a three-set Venn diagram

A group of 50 people are asked which sports they watch: football, golf and hockey.

A completed three-set Venn diagram showing the centre-first method for football, golf and hockey.

6 watch all three sports.
10 watch football and golf.
9 watch golf and hockey.
8 watch football and hockey.
28 watch football, 18 watch golf and 20 watch hockey.

Complete the Venn diagram and find the probability that two people chosen at random both watch football.

Put the “all three” value in the centre:
$6$
Subtract the centre from each pairwise total:
$only=8−6=2\begin{aligned} \text{football and golf only} &= 10-6=4\\ \text{golf and hockey only} &= 9-6=3\\ \text{football and hockey only} &= 8-6=2 \end{aligned}$
Find each “only” region:
$only=20−2−3−6=9\begin{aligned} \text{football only} &= 28-4-2-6=16\\ \text{golf only} &= 18-4-3-6=5\\ \text{hockey only} &= 20-2-3-6=9 \end{aligned}$
Find how many people watch none of the three sports:
$50 - (16 + 5 + 9 + 4 + 3 + 2 + 6) = 5$
For two people both watching football, sample without replacement:
$2850×2749=3781225≈0.309\frac{28}{50}\times\frac{27}{49}=\frac{378}{1225}\approx 0.309$

Independent events

Definition

Independent events

Two events are independent if the outcome of one event does not affect the probability of the other.

For independent events:

\cap B)=P(A)P(B)

This formula is the test you should use when asked whether two events are statistically independent.

Example

Testing for independence

A completed Venn diagram has these regions: $A$ only is 0.28, $B$ only is 0.18, the overlap is 0.12, and outside both is 0.42. Decide whether $A$ and $B$ are independent.

The Venn diagram for testing independence by comparing the overlap with the product of the event probabilities.

Find $P (A)$ by adding the regions inside circle $A$ :
$P (A) = 0.28 + 0.12 = 0.40$
Find $P (B)$ by adding the regions inside circle $B$ :
$P (B) = 0.18 + 0.12 = 0.30$
Compare the actual overlap with the product:
$P(A)P(B)=0.40×0.30=0.12P(A)P(B)=0.40\times 0.30=0.12$
Since this equals the overlap, the events are independent:
$\cap B)=P(A)P(B)$

Tip

How to write the reason

For independence questions, write a full reason such as: “Yes, because $\cap B)=P(A)P(B)$ .” A bare “yes” is not enough.

Tree diagrams

A tree diagram shows a probability experiment in stages. Each branch has a probability.

For staged events:

Multiply along branches for “and”.
Add separate routes for “or”.
If there is no replacement, update the numbers after the first choice.

Probability tree diagram for two counters drawn without replacement

Definition

Conditional probability

A conditional probability is the probability of an event given that something else has already happened.

It is written as $P(B∣A)P(B\mid A)$ , meaning “the probability of $B$ given $A$ ”.

Example

Two items without replacement

A two-stage tree diagram shows how probabilities change after the first chocolate is taken without replacement.

For two dark chocolates, multiply down the dark-then-dark route:
$P(DD)=818×717=28153P(DD)=\frac{8}{18}\times\frac{7}{17}=\frac{28}{153}$
For one of each type, there are two possible orders: milk then dark, or dark then milk.
$P(MD)=1018×817=40153P(DM)=818×1017=40153\begin{aligned} P(MD)&=\frac{10}{18}\times\frac{8}{17}=\frac{40}{153}\\ P(DM)&=\frac{8}{18}\times\frac{10}{17}=\frac{40}{153} \end{aligned}$
Add the two routes:
$each)=40153+40153=80153P(\text{one of each})=\frac{40}{153}+\frac{40}{153}=\frac{80}{153}$

Common Mistake

Forgetting the second order

“One of each” usually has two routes: first type A then type B, or first type B then type A. Work out both unless the question fixes the order.

Total probability

Sometimes an outcome can happen through several different routes. Find each route probability, then add them.

Example

Faulty components from different machines

A total probability tree shows the separate routes through machines A, B and C to a faulty component.

Convert the percentages to decimals before multiplying.
For “from A and faulty”, multiply along that route:
$faulty)=0.40×0.03=0.012P(\text{A and faulty})=0.40\times 0.03=0.012$
A faulty component can come from A, B or C, so add the three routes:
$P(faulty)=(0.40×0.03)+(0.35×0.02)+(0.25×0.01)P(\text{faulty})=(0.40\times 0.03)+(0.35\times 0.02)+(0.25\times 0.01)$
Calculate the final probability:
$P(faulty)=0.012+0.007+0.0025=0.0215P(\text{faulty})=0.012+0.007+0.0025=0.0215$

Exam technique

In the exam

Label every region or branch clearly before doing calculations.
For Venn diagrams, remember: all regions must add to 1 for probabilities, or to the total number of people/items for frequencies.
For tree diagrams, multiply along a route and add different routes; check whether the experiment is with or without replacement.

Self review

Check yourself

Can you explain the difference between mutually exclusive and independent events?
In a three-set Venn diagram, why do you fill the centre first?
When a question asks for “one of each”, have you considered both possible orders?

Probability

What you'll learn

Probability basics

Venn diagrams and set notation

Mutually exclusive events

Three-set Venn diagrams

Independent events

Tree diagrams

Total probability

Recap questions

You've reached the end

Probability

What you'll learn

Probability basics

Venn diagrams and set notation

Mutually exclusive events

Three-set Venn diagrams

Independent events

Tree diagrams

Total probability

Recap questions

You've reached the end

Probability

What you'll learn

Probability basics

Venn diagrams and set notation

Mutually exclusive events

Three-set Venn diagrams

Independent events

Tree diagrams

Total probability

Recap questions

You've reached the end

Probability Revision Guide

Probability

What you'll learn

Probability basics

Venn diagrams and set notation

Mutually exclusive events

Three-set Venn diagrams

Independent events

Tree diagrams

Total probability

Recap questions

You've reached the end

Probability Revision Guide