P6, P9Venn diagrams and set notation

Edexcel GCSE Maths Foundation (1MA1), Foundation tier · Probability

Practise Venn diagrams and set notation. 1 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.

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Revision notes

Venn diagrams sort items into sets and show how the sets overlap, and set notation (\(\xi\), \(\cap\), \(\cup\), \(A'\)) describes the regions. On both tiers you complete Venn diagrams from lists or totals and read probabilities from them. Harder questions use three sets or algebra, and on Higher you also find conditional probabilities ('given that') from a Venn diagram.

Grade by grade

What you need to be able to do, from the first marks up to the top grade.

3
Sort items into a two-set Venn diagramPut items in both sets in the overlap, the rest of each set in its own circle, and items in neither set outside the circles.
4
Understand ξ, intersection and union\(\xi\) is everything, \(A \cap B\) is the overlap, and \(A \cup B\) is everything inside either circle.
4
Find a probability from a Venn diagramDivide the number in the region you want by the total in \(\xi\), including anything outside the circles.
5
Complete a Venn diagram from given totalsStart with the overlap, subtract it from each set's total to get the 'only' regions, then work out the number outside.
5
Use complements such as A′ in set notation\(A'\) is everything not in \(A\), so \(A' \cap B\) is the part of \(B\) outside \(A\).

Notes

Sets and Venn diagrams

A set is a collection of items (elements), listed in curly brackets, e.g. \(A = \{2, 4, 6, 8\}\).
The universal set \(\xi\) is everything being considered. It is the rectangle round the diagram.
Each circle is one set. Items in both sets go in the overlap; items in neither set go inside the rectangle but outside the circles.
Every item goes in exactly one region. A Venn diagram can show the items themselves or just how many items are in each region.

Set notation

\(A \cap B\) (intersection): in \(A\) and in \(B\). This is the overlap.
\(A \cup B\) (union): in \(A\) or \(B\) or both. This is everything inside the circles.
\(A'\) (complement): not in \(A\). This includes the region outside both circles.
Combinations: \(A' \cap B\) is in \(B\) but not in \(A\) ('\(B\) only'), and \((A \cup B)'\) is outside both circles.

Filling in and finding probabilities

Start with the overlap. Then '\(A\) only' = total in \(A\) − overlap, and the same for \(B\). Then outside = grand total − everything in the circles.
Example: of 30 students, 18 study French, 12 study Spanish and 5 study both. French only = 13, Spanish only = 7, neither = 30 − 13 − 5 − 7 = 5.
P(region) = number in the region ÷ total number in \(\xi\). Here P(studies French or Spanish) = \(\frac{25}{30} = \frac{5}{6}\).
Check: all the numbers, including the one outside the circles, add up to the total.

Cheatsheet

\(\xi\) = the universal set (everything, the rectangle)
\(A \cap B\) = in \(A\) and in \(B\) (the overlap)
\(A \cup B\) = in \(A\) or \(B\) or both
\(A'\) = not in \(A\)
\((A \cup B)'\) = in neither \(A\) nor \(B\)
\(A\) only = total in \(A\) − number in \(A \cap B\)
P(region) = number in the region ÷ total in \(\xi\)

How to answer each type of question

Complete a Venn diagram from lists and use set notation

4 to 5 marks4

List the members of each set first.
Put members of both sets in the overlap, then the rest of each set, then everything else in \(\xi\) outside the circles.
Tick off each member of \(\xi\) as you place it, so nothing is missed or written twice.
For a probability, count the members in the region and divide by the number of members of \(\xi\).

Example. \(\xi\) = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
\(A\) = {multiples of 3}
\(B\) = {factors of 12}
(a) Complete a Venn diagram for this information. (2 marks)
(b) List the members of \(A \cap B\). (1 mark)
(c) A number is chosen at random from \(\xi\). Work out the probability that the number is in \(A \cup B\). (2 marks)

Show the model answer

(a) \(A\) = {3, 6, 9, 12} and \(B\) = {1, 2, 3, 4, 6, 12}.
Overlap: 3, 6, 12. \(A\) only: 9. \(B\) only: 1, 2, 4. Outside: 5, 7, 8, 10, 11 (B2 for a fully correct diagram, B1 if one or two numbers are misplaced)
(b) 3, 6, 12 (B1)
(c) \(A \cup B\) = {1, 2, 3, 4, 6, 9, 12}, which is 7 numbers (M1), so the probability is \(\frac{7}{12}\) (A1)

Complete a Venn diagram from totals and find a probability

4 to 5 marks5

Write the 'both' number in the overlap first.
Subtract it from each set's total to get the 'only' regions.
Subtract everything in the circles from the grand total to get the number outside.
Work out which region the notation describes (shade it if that helps), then divide by the grand total.

Example. 80 people at a gym were asked whether they use the pool and whether they use the sauna. 45 use the pool, 30 use the sauna and 12 use both. \(A\) is the set of people who use the pool and \(B\) is the set of people who use the sauna.
(a) Complete a Venn diagram. (3 marks)
(b) One of the 80 people is chosen at random. Work out the probability that this person is in \(A' \cap B\). (2 marks)

Show the model answer

(a) Overlap: 12 (B1)
Pool only: 45 − 12 = 33 and sauna only: 30 − 12 = 18 (B1)
Outside: 80 − 33 − 12 − 18 = 17 (B1)
(b) \(A' \cap B\) is 'sauna but not pool', which is 18 people (M1), so the probability is \(\frac{18}{80} = \frac{9}{40}\) (A1)

Shortcuts and memory tricks

\(\cup\) looks like a U, for Union: everything in either set. \(\cap\) is the other one: only the overlap, in both.
Work from the inside out: overlap first, then the 'only' regions, then the outside.
Shade the region the notation describes before you count. It stops you mixing up \(\cap\) and \(\cup\).
Check that every number in the diagram, including the outside, adds up to the total.

Where marks are lost

Putting the whole total for \(A\) in the '\(A\) only' region, so the overlap is counted twice.
Forgetting the number outside the circles when you find the total, or leaving that region blank.
Mixing up \(\cap\) (and: the overlap) with \(\cup\) (or: everything in the circles).
Thinking \(A'\) is just '\(B\) only': it also includes everything outside both circles.
Writing how many members a set has when the question says 'list the members'.
Higher: dividing by the grand total in a 'given that' question instead of by the number in the given set.

Exam technique

When you complete a Venn diagram of numbers, fill every region, including the outside. Write 0 if a region is empty.
'List the members' means write the actual elements, each one once.
For a probability, the denominator is the total in \(\xi\), unless the question says 'given that' (Higher), when it is the number in the given set.
In maths, 'or' includes both: 'plays football or tennis' means \(F \cup T\), which includes the people who play both.

Sample questions

Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.

Question 1Easy3 marks

The Venn diagram shows the sets ξ, \(A\) and \(B\).
ξ = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}

[object Object]

(a) List the members of the set \(A \cap B\).[1]

(b) One of the numbers in ξ is chosen at random.
Find \(P(B')\).[2]

Show the answer and mark scheme

(a) Answer: 3

B1 for 3 (and no others)

(b) Answer: \(\frac{3}{5}\) (or 0.6)

M1 for \(\frac{9}{15}\)
A1 for \(\frac{3}{5}\) oe

Worked solution: The members of \(B'\) are 1, 4, 6, 8, 9, 10, 12, 14, 15, so \(P(B') = \frac{9}{15} = \frac{3}{5}\).

Question 2Medium4 marks

There are 30 students in a class.
18 of the students play football, 15 play tennis and 5 play neither.

(a) Explain why some students must play both football and tennis.[1]

(b) A student is chosen at random from the class.
Work out the probability that this student plays tennis but not football.[3]

Show the answer and mark scheme

(a) Answer: \(18 + 15 + 5 = 38\), which is more than 30, so 8 students have been counted twice.

C1 for \(18 + 15 + 5 = 38 \gt 30\) (or \(18 + 15 = 33 \gt 25\)), so some students are in both groups

Worked solution: Only 25 students play at least one sport, but \(18 + 15 = 33\). The extra 8 must be students counted in both groups.

(b) Answer: \(\frac{7}{30}\)

M1 for both = \(38 - 30 = 8\)
M1 for \(15 - 8 = 7\) playing tennis but not football (ft their 8)
A1 for \(\frac{7}{30}\) oe

Worked solution: Both: \(33 - 25 = 8\). Tennis only: \(15 - 8 = 7\). Probability \(= \frac{7}{30}\).

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