Are and easy to confuse?

Remember: looks like a cup that holds everything from both sets (union); is the narrow shared part (intersection).

P6, P9ProbabilityFoundation & HigherGrades 4–7

Venn Diagrams

A Venn diagram sorts things into overlapping groups. Each region holds the items that belong to exactly that combination of sets, so probabilities can be read straight off the diagram. Venn diagrams also make conditional probability — "given that" questions — easy to see.

45 min Video by GCSE Maths AQA GCSE Maths

The big picture

Always fill a Venn diagram from the $\textbf{middle outwards}$ . The overlap is counted inside both circles, so if 12 people like tea, 5 of whom also like coffee, only 7 go in the "tea only" region.

Set notation is a compact language: $A \cap B$ ("A and B", the overlap), $A \cup B$ ("A or B", anything in either circle), $A'$ ("not A", everything outside $A$ ), and $\xi$ (the universal set — everything).

Probability using Venn DiagramsCurated by Kepler, not produced by us. Watch it, then read the notes below.

Open on YouTube

What you'll learn

Read and use set notation: $\xi$ , $\cap$ , $\cup$ , $A'$
Complete Venn diagrams from given information, including with algebra
Calculate probabilities from Venn diagrams
Calculate conditional probabilities from a Venn diagram (Higher)

Set notation

$\xi$ is the universal set: everything being considered. $A \cap B$ is the $\textbf{intersection}$ — in both $A$ and $B$ . $A \cup B$ is the $\textbf{union}$ — in $A$ or $B$ or both. $A'$ is the $\textbf{complement}$ — not in $A$ .

Combinations work too: $A \cap B'$ means "in $A$ but not in $B$ ".

30 students: 15 study art (

A

), 17 study biology (

B

), 6 study both, 4 study neither.

Example 1Using the diagram, find

n(A \cup B)

n(A')

and

n(A \cap B')

A \cup B

: everything inside a circle:

9 + 6 + 11 = 26

A'

: everything outside

A

11 + 4 = 15

A \cap B'

: in

A

but not

B

9

Answer

26

15

9

Your turnSame method, new numbers — try it before moving on.

Using the diagram, how many students are in $A' \cap B'$ ?

Your answer

Show worked solution

Solution

1Not in $A$ and not in $B$ : outside both circles.
2That region contains 4.

Answer: $4$

Completing Venn diagrams

Start with the intersection. Subtract it from each set's total to find the "only" regions. Finally, subtract everything inside the circles from the universal total to find the outside.

If a region is unknown, call it $x$ and form an equation from the total.

Example 1In a group of 50 people, 28 have a cat, 20 have a dog and 9 have both. How many have neither?

1Both: 9. Cat only:

28 - 9 = 19

. Dog only:

20 - 9 = 11

2Inside circles:

19 + 9 + 11 = 39

3Neither:

50 - 39 = 11

Answer

11

people

Your turnSame method, new numbers — try it before moving on.

40 students were asked about football (F) and tennis (T). 25 play football, 12 play tennis and 8 play neither. How many play both?

Your answer

Show hint

Let both $= x$ . Then football only is $25 - x$ and tennis only is $12 - x$ .

Show worked solution

Solution

1 $(25 - x) + x + (12 - x) + 8 = 40$ .
2 $45 - x = 40$ , so $x = 5$ .

Answer: $5$

Tip — Check that all four regions add up to the universal total.

Probabilities from Venn diagrams

Probability $= \dfrac{\text{number in the region}}{\text{total in } \xi}$ . If the diagram already shows probabilities, read or add them directly.

Example 1Using the art and biology diagram, a student is chosen at random. Find

P(A \cup B)

and

P(B \cap A')

P(A \cup B) = \frac{26}{30} = \frac{13}{15}

P(B \cap A') = \frac{11}{30}

Answer

\frac{13}{15}

and

\frac{11}{30}

Your turnSame method, new numbers — try it before moving on.

Using the art and biology diagram, find $P(A')$ .

Your answer

Show worked solution

Solution

1Not in $A$ : $11 + 4 = 15$ students.
2 $P(A') = \frac{15}{30} = \frac{1}{2}$ .

Answer: $\frac{1}{2}$

Conditional probability (Higher)

"Given that" restricts you to part of the diagram. $P(A \mid B)$ — the probability of $A$ given $B$ — only looks inside circle $B$ : $P(A \mid B) = \dfrac{n(A \cap B)}{n(B)}$ .

Example 1Using the art and biology diagram, a student studies biology. Find the probability that they also study art.

1Given biology: only the 17 students in

B

count.

2Of those, 6 also study art.

Answer

P(A \mid B) = \frac{6}{17}

Your turnSame method, new numbers — try it before moving on.

Using the art and biology diagram, a student who does NOT study biology is chosen. What is the probability they study art?

Your answer

Show hint

Not biology: the regions outside $B$ are "art only" (9) and "neither" (4).

Show worked solution

Solution

1Outside $B$ : $9 + 4 = 13$ students.
2Of those, 9 study art: $\frac{9}{13}$ .

Answer: $\frac{9}{13}$

Changing the denominator

The "given" condition changes the denominator, not the numerator region. $P(A \cap B) = \frac{6}{30}$ looks at everyone; $P(A \mid B) = \frac{6}{17}$ looks only at biology students. Mixing up these two is the most common Higher-tier probability error.

Think like an examiner

Fill in the intersection first and show the subtractions — completed diagrams often earn marks before any probability is found.
Include the outside region (inside $\xi$ but outside the circles) — it is frequently missed.
For "given that", write the fraction with the restricted total as the denominator.

Set notation

A \cap B, \quad A \cup B, \quad A'

And (overlap), or (either), not.

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

Do not count the overlap twice.

P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}

Conditional probability.

Watch out for these

Writing the set total (e.g. 28) in the "only" region.

Subtract the overlap first:

28 - 9 = 19

Treating

A \cup B

as just the overlap.

\cup

is everything in either circle;

\cap

is the overlap.

P(A \mid B) = \frac{n(A \cap B)}{\text{total}}

Divide by

n(B)

Forgetting the region outside both circles.

Subtract the circle total from

\xi

Stretch yourself

In a survey of 80 people, twice as many like crisps as like popcorn. 12 like both, and 14 like neither. How many like popcorn?

Hint — Let the number who like popcorn be $p$ ; then $2p$ like crisps. Use $n(A \cup B) = n(A) + n(B) - n(A \cap B)$ .

Reveal solution

SolutionInside the circles:

80 - 14 = 66

. So

2p + p - 12 = 66

, giving

3p = 78

and

p = 26

. So 26 like popcorn (and 52 like crisps). Check regions: popcorn only 14, both 12, crisps only 40, neither 14: total 80 ✓.

\blacksquare

Questions students ask

What does

\xi

mean?

The universal set — everything under consideration. The rectangle around a Venn diagram represents it.

Can Venn diagrams have three circles?

Yes. Fill them from the centre (the triple overlap) outwards in the same way.

Are

\cap

and

\cup

easy to confuse?

Remember:

\cup

looks like a cup that holds everything from both sets (union);

\cap

is the narrow shared part (intersection).

Key takeaways

$\cap$ = and (intersection); $\cup$ = or (union); $A'$ = not $A$ .
Fill Venn diagrams from the intersection outwards.
Use algebra for unknown regions and the universal total.
Probability = region ÷ total.
Given that: restrict the denominator to the given set.

Exam-style questions

Practise it the way it is examined

2 original questions · 6 marks, written to match the style and mark allocation of the real papers. Work on paper, then open the mark scheme and tick the marks you earned — method marks count even if the final answer slips.

Question 1Grade 43 marks

$\xi = \{1, 2, 3, \ldots, 12\}$ . $A$ is the set of even numbers. $B$ is the set of multiples of 3.

(a)List the members of $A \cap B$ .[1]
Show mark scheme (a)
- B1 $6, 12$
(b)A number is chosen at random from $\xi$ . Find $P(A \cup B)$ .[2]
Show mark scheme (b)
- M1Identifies $A \cup B = \{2, 3, 4, 6, 8, 9, 10, 12\}$
- A1 $\frac{8}{12} = \frac{2}{3}$
Answer: $\frac{2}{3}$

Question 2Grade 63 marks

60 people were asked if they have been to France ( $F$ ) or Spain ( $S$ ). 31 have been to France. 24 have been to Spain. 15 have been to neither.

(a)Complete a Venn diagram to show this information.[2]
Show mark scheme (a)
- M1Finds the intersection: $31 + 24 + 15 - 60 = 10$
- A1Regions 21, 10, 14, 15 correctly placed
(b)One of the people who has been to Spain is chosen at random. Find the probability they have also been to France.[1]
Show mark scheme (b)
- B1 $\frac{10}{24} = \frac{5}{12}$ (follow through)
Answer: $\frac{5}{12}$

Unlimited practice

Skill drill

A new question every time, marked instantly, with a full worked solution. Aim for a streak of five, then move up a level.

0 / 0 first-time correct Streak 0Best at this level: 0

Generating a question…

Test yourself

Answer a few questions, then review anything you missed.

Prefer paper? Open the written self-check

Answer each question before opening its answer. This self-check does not record a quiz score.

$A \cap B$ means…
- in both A and B
- in A or B
- not in A
- in A but not B
Show answer and explanation
Answer: in both A and B
The intersection.
$A \cup B$ means…
- in A or B or both
- only in both
- in neither
- not in B
Show answer and explanation
Answer: in A or B or both
The union.
$n(A) = 15$ and $n(A \cap B) = 4$ . $n(A \text{ only})$ ?
- $11$
- $19$
- $15$
- $4$
Show answer and explanation
Answer: $11$
$15 - 4$ .
$n(\xi) = 40$ , $n(A \cup B) = 33$ . How many are in neither?
- $7$
- $33$
- $73$
- $40$
Show answer and explanation
Answer: $7$
$40 - 33$ .
$n(A) = 20$ , $n(B) = 15$ , $n(A \cap B) = 6$ . $n(A \cup B)$ ?
- $29$
- $35$
- $41$
- $26$
Show answer and explanation
Answer: $29$
$20 + 15 - 6$ .
Regions: A only 8, both 4, B only 6, neither 2. $P(B)$ ?
- $\frac{1}{2}$
- $\frac{3}{10}$
- $\frac{2}{5}$
- $\frac{1}{5}$
Show answer and explanation
Answer: $\frac{1}{2}$
$\frac{10}{20}$ .
Same diagram: $P(A \mid B)$ ?
- $\frac{2}{5}$
- $\frac{1}{5}$
- $\frac{1}{3}$
- $\frac{4}{12}$
Show answer and explanation
Answer: $\frac{2}{5}$
$\frac{4}{10}$ .
Same diagram: $P(A' \cap B)$ ?
- $\frac{3}{10}$
- $\frac{1}{2}$
- $\frac{2}{5}$
- $\frac{1}{10}$
Show answer and explanation
Answer: $\frac{3}{10}$
B only: $\frac{6}{20}$ .

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