Venn diagrams & set notation – GCSE Maths exam questions with mark scheme

1Grade 1–3FoundationNon-calculator3 marks

$\xi=\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\}$

$A=$ even numbers $B=$ multiples of 3

(a) Complete the Venn diagram. [2]

A number is chosen at random from $\xi$.

(b) Find the probability that the number is in $A \cap B$. [1]

(a) 6 and 12 in the intersection onlyB1
(a) 2, 4, 8, 10 in A only, 3, 9 in B only and 1, 5, 7, 11 outside both circles (all 12 numbers placed once)B1
(b) $\frac{2}{12}$ oe (ft their Venn diagram)B1
Answer: (a) $A$ only: 2, 4, 8, 10; $A\cap B$: 6, 12; $B$ only: 3, 9; outside: 1, 5, 7, 11 (b) $\frac{2}{12}=\frac{1}{6}$
Examiner tip: Numbers in neither set still go in the diagram — inside the rectangle but outside both circles.

2Grade 4–5FoundationNon-calculator3 marks

There are 50 students in a club.

28 students play football. 17 students play tennis. 8 students play both football and tennis.

(a) Work out how many students play neither football nor tennis. You may use the Venn diagram to help you. [2]

(b) A student is chosen at random. Find the probability that the student plays tennis but not football. [1]

(a) $28+17-8$ $(=37)$ or $20+8+9$, or 20, 8 and 9 in the correct regions of the Venn diagramM1
(a) 13 caoA1
(b) $\frac{9}{50}$ oe (accept 0.18 or 18%; ft their tennis-only value $\div 50$)B1
Answer: (a) 13 (b) $\frac{9}{50}$
Examiner tip: The 8 students who play both are already counted in the 28 and in the 17 — subtract them once so they are not counted twice.

3Grade 4–5Non-calculator3 marks

The Venn diagram shows information about a group of 40 people.

$A$ = people who own a bike $B$ = people who own a car

One of the 40 people is chosen at random.

(a) Find $P(A \cup B)$. [1]

(b) Find $P(A')$. [1]

(a) $\frac{34}{40}$ oe (e.g. $\frac{17}{20}$, 0.85)B1
(b) $\frac{21}{40}$ oe (e.g. 0.525)B1
(c) 12 (reject $\frac{12}{40}$)B1
Answer: (a) $\frac{34}{40}$ (b) $\frac{21}{40}$ (c) 12
Examiner tip: $A'$ means NOT in $A$ — include the 6 people outside both circles as well as the 15 in $B$ only.

4Grade 4–5HigherNon-calculator3 marks

The Venn diagram shows the probabilities for events $A$ and $B$.

(a) Work out the value of $x$. [2]

(b) Work out $P(B')$. [1]

(a) $0.3+x+2x+0.16=1$ or $3x=0.54$M1
(a) 0.18A1
(b) 0.46 oe (ft $1-3\times$ their $x$, or $0.3+0.16$)B1
Answer: (a) $x=0.18$ (b) 0.46
Examiner tip: $P(B')$ includes the $A$-only region and the outside region: $0.3+0.16=0.46$ is a useful check.

5Grade 6–7HigherNon-calculator3 marks

100 people were asked if they own a cat ($C$), a dog ($D$) or a rabbit ($R$). The Venn diagram shows the results.

(a) How many of the people own a cat but not a dog? [1]

(b) One of the people is chosen at random. Work out the probability that this person owns exactly two of these types of pet. [2]

(a) 26B1
(b) $9+4+6$ $(=19)$M1
(b) $\frac{19}{100}$ oeA1
Answer: (a) 26 (b) $\frac{19}{100}$
Examiner tip: 'Exactly two' excludes the centre region — the 3 people who own all three pets.

6Grade 6–7HigherNon-calculator5 marks

There are 32 students in a class. Each student is asked if they study French and if they study Spanish.

$x$ students study both French and Spanish.
Twice as many students study French only as study both.
5 more students study Spanish only than study both.
3 students study neither language.

You may use the Venn diagram to help you.

A student is chosen at random. Work out the probability that this student studies Spanish. [5]

$2x$ for French only and $x+5$ for Spanish only (may be on a Venn diagram)M1
$2x+x+x+5+3=32$ oeM1
$x=6$A1
$\frac{x+(x+5)}{32}$ using their $x$M1
$\frac{17}{32}$ oe (accept 0.53125 or 0.531)A1
Answer: $\frac{17}{32}$
Examiner tip: 'Studies Spanish' means the whole Spanish circle — include the students who study both, not just Spanish only.

7Grade 6–7HigherNon-calculator2 marks

Use set notation to describe the shaded region in each Venn diagram.

(a) [1]

(b) [1]

(a) $A \cap B'$ (accept equivalent, e.g. $A \cap (A\cap B)'$)B1
(b) $(A \cup B)'$ (accept $A' \cap B'$)B1
Answer: (a) $A \cap B'$ (b) $(A \cup B)'$
Examiner tip: $A' \cup B'$ is NOT the same as $(A\cup B)'$ — it includes the $A$-only and $B$-only regions too.

8Grade 1–3FoundationNon-calculator3 marks

$\xi=\{30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40\}$

$A=$ multiples of 4 $B=$ multiples of 6

The Venn diagram shows this information.

(a) Write down all the numbers that are in set $A$ but not in set $B$. [1]

(b) A number is chosen at random from $\xi$. Find the probability that the number is not in set $B$. [1]

(c) Kai says, “48 is a multiple of 4 and a multiple of 6, so 48 should be written in the middle of the Venn diagram.”
Explain why Kai is wrong. [1]

(a) 32 and 40 (both, and no others); ignore orderB1
(b) $\frac{9}{11}$ oe (accept 0.818… or 81.8%); do not accept ratio 9 : 11 or ‘9 out of 11’ in words aloneB1
(c) 48 is not in $\xi$ oe, e.g. ‘the universal set only goes from 30 to 40’; do not accept ‘48 is too big’ without reference to $\xi$ / the numbers in the setC1
Answer: (a) 32, 40 (b) $\frac{9}{11}$ (c) 48 is not in the universal set $\xi$ (which only contains 30 to 40), so it cannot go anywhere on the diagram.
Examiner tip: ‘Not in B’ includes the numbers outside both circles — count all 9, not just the 2 in A only.

9Grade 4–5FoundationNon-calculator4 marks

60 people went to a cinema.

35 of them bought a drink.
12 of them bought popcorn and a drink.
9 of them bought neither popcorn nor a drink.

$P$ = people who bought popcorn $D$ = people who bought a drink

(a) Complete the Venn diagram. [3]

(b) One of the 60 people is chosen at random. Find the probability that this person bought popcorn. [1]

(a) 12 in the intersection $P \cap D$B1
(a) 23 in $D$ only (35 − 12)B1
(a) 16 in $P$ only and 9 outside both circles (ft 60 − 9 − their total for $D$, provided all four numbers sum to 60)B1
(b) $\frac{28}{60}$ oe, e.g. $\frac{7}{15}$, 0.466… (ft their $P$ only + 12 over 60)B1
Answer: (a) $P$ only 16, $P\cap D$ 12, $D$ only 23, outside 9 (b) $\frac{28}{60}=\frac{7}{15}$
Examiner tip: 35 is the total for the whole $D$ circle, so the ‘drink only’ part is 35 − 12 = 23, not 35.

10Grade 4–5HigherNon-calculator3 marks

(a) On the first Venn diagram, shade the region $A \cup B'$. [1]

(b) On the second Venn diagram, shade the region $(A \cap B)'$. [1]

(c) Sam says, “$(A \cup B)'$ and $A' \cap B'$ are the same region.”
Is Sam correct? Give a reason for your answer. [1]

(a) Shades $A$ (all of it, including the overlap) and the region outside both circles; $B$-only region left unshadedB1
(b) Shades everything except the overlap: $A$ only, $B$ only and the region outside both circlesB1
(c) Yes, with a correct reason, e.g. ‘both are the region outside both circles’ / ‘not in A and not in B is the same as not in A or B’; accept a correct pair of sketches as the reasonC1
Answer: (a) everything except ‘$B$ only’ (b) everything except the overlap $A\cap B$ (c) Yes — both describe the region outside both circles (in neither $A$ nor $B$).
Examiner tip: $A \cup B'$ includes the area outside both circles — students often shade only circle $A$ and drop the mark.

11Grade 6–7HigherNon-calculator5 marks

$\xi=\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15\}$

$A=$ even numbers $B=$ factors of 12 $C=$ prime numbers

(a) Complete the Venn diagram. [3]

(b) List the members of the set $B \cap C'$. [1]

A number is chosen at random from $\xi$.

(a) at least 7 of the 15 numbers placed correctly (each number appearing once)B1
(a) at least 11 of the 15 numbers placed correctly (each number appearing once)B1
(a) all 15 numbers placed correctly: centre 2; $A\cap B$ only 4, 6, 12; $B\cap C$ only 3; $A\cap C$ only empty; $A$ only 8, 10, 14; $B$ only 1; $C$ only 5, 7, 11, 13; outside 9, 15B1
(b) 1, 4, 6, 12 (ft their diagram); ignore brackets/orderB1
(c) $\frac{6}{15}$ oe, e.g. $\frac{2}{5}$ (ft their diagram)B1
Answer: (a) centre 2; $A\cap B$ only 4, 6, 12; $B\cap C$ only 3; $A\cap C$ only none; $A$ only 8, 10, 14; $B$ only 1; $C$ only 5, 7, 11, 13; outside 9, 15 (b) $\{1, 4, 6, 12\}$ (c) $\frac{6}{15}=\frac{2}{5}$
Examiner tip: 1 is a factor of 12 but not a prime number — it goes in $B$ only, a very common misplacement.

12Grade 6–7HigherNon-calculator4 marks

The Venn diagram gives information about the GCSE options of 80 students.

$G$ = students who study Geography $H$ = students who study History

(a) A student is chosen at random. Given that the student studies History, work out the probability that the student also studies Geography. [2]

(b) Work out $P(H \mid G')$. [2]

(a) Uses 36 (= 15 + 21) as the denominator, e.g. $\frac{\ldots}{36}$M1
(a) $\frac{15}{36}$ oe, e.g. $\frac{5}{12}$A1
(b) Uses 38 (= 21 + 17) as the denominator, e.g. $\frac{\ldots}{38}$M1
(b) $\frac{21}{38}$ oe (do not accept $\frac{21}{80}$)A1
Answer: (a) $\frac{15}{36}=\frac{5}{12}$ (b) $\frac{21}{38}$
Examiner tip: ‘Given that’ shrinks the denominator to that group only — dividing by 80 scores zero.

13Grade 6–7HigherCalculator4 marks

The Venn diagram shows the probabilities of two events, $A$ and $B$.

(a) Show that events $A$ and $B$ are not independent. [3]

(b) Work out $P(B \mid A)$. [1]

(a) $P(A)=0.4$ and $P(B)=0.44$ (both found)M1
(a) $P(A)\times P(B)=0.4\times0.44$ $(=0.176)$; alt: $P(A\mid B)=\frac{0.12}{0.44}$ $(=0.27…)$ or $P(B\mid A)=\frac{0.12}{0.4}$ $(=0.3)$M1
(a) Compares with a correct value: $0.176 \neq 0.12 = P(A\cap B)$ (or $0.27… \neq 0.4$, or $0.3 \neq 0.44$) so not independentC1
(b) 0.3 oe, e.g. $\frac{3}{10}$B1
Answer: (a) $P(A)P(B)=0.4\times0.44=0.176\neq0.12=P(A\cap B)$, so not independent (b) 0.3
Examiner tip: Use the whole circle for $P(A)$ (0.28 + 0.12), not just ‘A only’, before multiplying.

14Grade 8–9HigherCalculator5 marks

A club has 25 members.

$C$ = members who play chess $D$ = members who take part in debating

The Venn diagram shows the number of members in each region.

Two different members are chosen at random. The probability that both of them are in $D$ only (debating but not chess) is $\frac{7}{100}$.

(a) Show that $x^2 - x - 42 = 0$ [2]

(b) Work out the probability that one of the two members chosen is in $C$ only and the other is in $D$ only. [3]

(a) $\frac{x}{25}\times\frac{x-1}{24}=\frac{7}{100}$ oeM1
(a) Correct algebra to $x^2-x-42=0$ with no errors, e.g. $100x(x-1)=4200 \Rightarrow x^2-x=42$A1
(b) Solves to get $x=7$ (rejecting $x=-6$; may be implied by use of 7)P1
(b) $\frac{8}{25}\times\frac{7}{24}$ $(=\frac{56}{600})$ and recognises two orders, e.g. $\times 2$ or $+\frac{7}{25}\times\frac{8}{24}$M1
(b) $\frac{112}{600}$ oe, e.g. $\frac{14}{75}$ or 0.1866…(0.187)A1
Answer: (a) $\frac{x(x-1)}{600}=\frac{7}{100}\Rightarrow x^2-x-42=0$ (b) $x=7$; $2\times\frac{8}{25}\times\frac{7}{24}=\frac{14}{75}$
Examiner tip: The second member is chosen from 24 people, not 25 — and the two members can be picked in either order, so double it.

15Grade 8–9HigherNon-calculator5 marks

Some students were asked which of three sports they play: badminton ($B$), hockey ($H$) and netball ($N$).

The Venn diagram shows the numbers of students. $x$ is a positive integer.

Given that a student plays netball, the probability that the student also plays hockey is $\frac{7}{20}$.

(a) Work out the value of $x$. [3]

(b) One of the students who does not play hockey is chosen at random. Work out the probability that this student plays badminton. [2]

(a) Expression for $n(N)$: $2x+5+x+3$ $(=3x+8)$ or for $n(H\cap N)$: $x+3$M1
(a) Forms a correct equation, e.g. $\frac{x+3}{3x+8}=\frac{7}{20}$M1
(a) $x=4$ caoA1
(b) Denominator $14+5+2x+9$ $(=36$ with $x=4)$ (ft their $x$)M1
(b) $\frac{19}{36}$ (ft their $x$: $\frac{19}{28+2x}$)A1
Answer: (a) $x=4$ (b) $\frac{19}{36}$
Examiner tip: For ‘does not play hockey’ include the students outside all three circles (9) in the denominator.

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