Venn Diagrams and Two-Way Tables · Study Notes · Edexcel IGCSE 4MA1 Higher

Venn diagrams for sets/probability; two-way tables for cross-categorising. Conditional probability from both is heavily examined.

At a glance

$\cap$ = intersection (AND).
$\cup$ = union (OR).
$A'$ = complement (NOT in $A$ ).
$\xi$ = universal set (everything).
Fill Venn from intersection OUTWARDS.
$|A \cup B| = |A| + |B| - |A \cap B|$ .
Conditional: $P(A|B) = P(A \cap B) / P(B)$ .
Two-way table: rows × columns, with totals.

What you’ll learn

Mapped to the Pearson Edexcel IGCSE 4MA1 syllabus (2026 onwards).

6.6 — Use set notation: $\cap, \cup, ', \xi$ .
6.6 — Construct and interpret Venn diagrams (2 and 3 sets).
6.6 — Calculate probabilities using Venn diagrams.
6.6 — Construct and interpret two-way tables.
6.6 — Calculate conditional probability from a Venn diagram or two-way table.

Set notation

▼

Master the four symbols: $\cap, \cup, ', \xi$ .

Set — a collection of distinct objects.

Universal set $\xi$ — everything under consideration. Drawn as a rectangle.

$A \cap B$ (A intersection B) — elements in BOTH $A$ AND $B$ . The overlap.

$A \cup B$ (A union B) — elements in $A$ OR $B$ (or both).

$A'$ (complement of $A$ ) — elements in $\xi$ but NOT in $A$ .

Memory aid for $\cap$ vs $\cup$ :

$\cap$ — pointed top (like the 'A' in AND). Intersection.
$\cup$ — opens upward like a cup. Union.

Shaded regions: intersection (the overlap), union (either circle), complement (everything outside A).

Examples. $\xi = \{1, 2, 3, 4, 5, 6, 7, 8\}$ , $A = \{2, 4, 6\}$ , $B = \{4, 5, 6, 7\}$ .

$A \cap B = \{4, 6\}$ .
$A \cup B = \{2, 4, 5, 6, 7\}$ .
$A' = \{1, 3, 5, 7, 8\}$ (not in $A$ ).
$(A \cup B)' = \{1, 3, 8\}$ (not in either).

Edexcel tip. Set questions sometimes use words: "elements in $A$ but not in $B$ " = $A \cap B'$ = $A$ minus the overlap.

$\cap$ = AND (intersection).
$\cup$ = OR (union).
$A'$ = NOT $A$ (complement).
$\xi$ = universal (everything).

Venn diagrams (2 and 3 sets)

▼

Fill from the INTERSECTION outwards.

Two-set Venn. Two overlapping circles inside a rectangle ( $\xi$ ). Four regions:

$A$ only (left).
$A \cap B$ (overlap).
$B$ only (right).
Outside both (corner of rectangle).

Filling rule. Always start with the intersection.

Worked example. $|A| = 12$ , $|B| = 15$ , $|A \cap B| = 7$ , total = 25.

$|A \cap B| = 7$ in the overlap.
$A$ only $= 12 - 7 = 5$ .
$B$ only $= 15 - 7 = 8$ .
Outside $= 25 - (5 + 7 + 8) = 5$ .

Sanity check: $5 + 7 + 8 + 5 = 25$ . ✓

Fill the overlap first, then subtract to get the 'only' regions; the rest sits outside.

Three-set Venn. Three overlapping circles, EIGHT regions. Same rule: start with the central intersection $A \cap B \cap C$ , work outwards.

Inclusion-exclusion (3 sets). $|A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |A \cap C| - |B \cap C| + |A \cap B \cap C|$

(Add singles, subtract pairs, add the triple — to correct for double/triple counting.)

Edexcel tip. Mark schemes for filling Venns:

M1: intersection placed correctly.
M1: 'only' regions correct.
A1: outside region correct.
B1: total checks out.

Start with intersection.
$A$ only = $|A|$ − intersection.
Outside = total − (sum of regions).
Always check sum equals total.

See the full worked example for venn diagrams and tables →

Probability from a Venn diagram

▼

Count the favourable region; divide by total.

Once a Venn is filled, probability questions are direct.

Worked example (continuing). 25 students. $|A \cap B| = 7$ , $A$ only $= 5$ , $B$ only $= 8$ , outside $= 5$ .

$P(A) = 12/25$ (size of $A$ over total).
$P(A \cap B) = 7/25$ .
$P(A \cup B) = 20/25 = 4/5$ .
$P(\text{neither}) = 5/25 = 1/5$ .
$P(\text{exactly one}) = (5 + 8)/25 = 13/25$ .

Conditional probability from Venn. $P(A | B)$ = "probability $A$ given $B$ ".

Restrict attention to $B$ only.
Of those, find how many are in $A$ .

$P(A | B) = \dfrac{P(A \cap B)}{P(B)} = \dfrac{|A \cap B|}{|B|}$

Worked example. $P(A | B) = 7/15$ (7 in both, 15 in $B$ in total).

Edexcel tip. When you see "given", "if it is known", "out of those who" — that's a conditioning signal. The denominator becomes the conditioning set.

Count favourable region.
Denominator = total (or conditioning set).
$P(A|B) = |A \cap B| / |B|$ .
Conditional changes the denominator.

See the full worked example for venn diagrams and tables →

Two-way tables

▼

Cross-categorise two variables. Row totals + column totals.

A two-way table sorts data by two categorical variables — one across rows, one across columns.

Example.

	Walks	Bus	Car	Total
Male	12	8	6	26
Female	14	12	8	34
Total	26	20	14	60

Each cell = number with that combination. Totals row + column show one-variable counts.

Inner cells hold the counts; the shaded row and column are totals, meeting at the grand total.

Reading probabilities.

$P(\text{walks AND male}) = 12/60 = 1/5$ .
$P(\text{walks}) = 26/60 = 13/30$ .
$P(\text{walks} | \text{male}) = 12/26 = 6/13$ (restricted to male row, total 26).

Filling missing cells. Use row + column totals to back-fill.

Worked example. Below, missing cells marked $?$ .

	A	B	Total
X	?	5	12
Y	7	?	18
Total	?	?	30

Row X total = 12, B-X = 5 → A-X = 7.
Row Y total = 18, A-Y = 7 → B-Y = 11.
Col A total = A-X + A-Y = 7 + 7 = 14.
Col B total = 5 + 11 = 16. Check: 14 + 16 = 30. ✓

Edexcel tip. Rows + columns + totals must all be consistent. A two-way table problem reduces to systematic addition/subtraction.

Two categories cross-tabulated.
Row + column totals give one-variable counts.
Conditional: restrict to row or column.
Back-fill via subtraction.

Quick recap

Set notation: $\cap$ (AND), $\cup$ (OR), $'$ (NOT), $\xi$ (universal).
Fill Venn from INTERSECTION outwards.
$|A \cup B| = |A| + |B| - |A \cap B|$ .
Conditional: $P(A|B) = P(A \cap B) / P(B)$ — denominator is the conditioning set.
Two-way tables: rows × columns + totals.
Always check totals sum correctly.

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

$A \cap B$ = AND (overlap).
$A \cup B$ = OR (everything in either).
$A'$ = NOT in $A$ .
$P(A|B) = P(A \cap B) / P(B)$ .
Inclusion-exclusion: $|A \cup B| = |A| + |B| - |A \cap B|$ .

How it’s examined

Venns and two-way tables are 4-6 mark questions on both papers. Conditional probability from these is the most common Higher Tier extension. Examiner reports flag: (1) putting $|A|$ in the 'only $A$ ' region, (2) wrong denominator for conditional, (3) confusing $\cap$ with $\cup$ .

Sources: Pearson Edexcel International GCSE Mathematics A 4MA1 specification (2026 onwards); 4MA1 Examiner Reports — Jan and May/Jun 2022-2024 series; Past Papers — 4MA1/1H May/Jun 2024, 4MA1/2H Jan 2024. Last reviewed 2026-05-07.

Step-by-step solutions to past-paper-style questions on venn diagrams and tables, written exactly the way a tutor would explain them at the board.

1Fill a Venn diagram
Higher• venn
▼
Question
$|A| = 12$ , $|B| = 15$ , $|A \cap B| = 7$ , total = 25. Fill the Venn diagram.
Step-by-step solution
1. Step 1
  Start with the intersection: $|A \cap B| = 7$ .
2. Step 2
  $A$ only = $|A| - |A \cap B| = 12 - 7 = 5$ .
3. Step 3
  $B$ only = $15 - 7 = 8$ .
4. Step 4
  Outside = total − (5 + 7 + 8) = $25 - 20 = 5$ .
Answer
$A$ only: 5; intersection: 7; $B$ only: 8; outside: 5. Total = 25.
Examiner tip
ALWAYS start with the intersection. Mark scheme awards M1 for correctly placed intersection.
2Probability from a Venn diagram
Higher• Adapted from 4MA1/1H May/Jun 2024 Q23• venn, probability
▼
Question
From the diagram above, find $P(A \cap B)$ , $P(A \cup B)$ , $P(\text{neither})$ .
Step-by-step solution
1. Step 1
  $P(A \cap B) = 7/25$ .
2. Step 2
  $P(A \cup B) = (5 + 7 + 8)/25 = 20/25 = 4/5$ .
3. Step 3
  $P(\text{neither}) = 5/25 = 1/5$ .
Answer
$P(A \cap B) = 7/25$ ; $P(A \cup B) = 4/5$ ; $P(\text{neither}) = 1/5$ .
3Conditional probability from Venn
Higher• conditional, venn
▼
Question
From the diagram, find $P(A | B)$ — probability $A$ given $B$ .
Step-by-step solution
1. Step 1
  Restrict to $B$ : there are 15 in $B$ in total.
2. Step 2
  Of those, 7 are also in $A$ . $P(A | B) = 7/15$ .
Answer
$P(A | B) = 7/15$
Examiner tip
Conditional: condition is the new total. $P(A|B)$ = (in both) / (in $B$ ).
4Two-way table
Higher• Adapted from 4MA1/2H Jan 2024 Q15• two-way table
▼
Question
Two-way table of students by gender and travel mode. Find $P(\text{walks AND male})$ .
Step-by-step solution
1. Step 1
  Read the cell: e.g., 12 students walk AND are male.
2. Step 2
  Total students = 60.
3. Step 3
  $P = 12/60 = 1/5$ .
Answer
$P(\text{walks AND male}) = 1/5$
5Probability of a set from Venn diagram counts [1 mark]
Getting started• Paper 1H/2H style — 1 mark• venn diagram, probability
▼
Question
In a group of $25$ people, $7$ own a bike only, $3$ own a bike and a scooter, $10$ own a scooter only and $5$ own neither.

One of the people is chosen at random. Write down the probability that the person owns a scooter. [1]
Step-by-step solution
1. Step 1
  Owning a scooter includes the people who own a scooter only and those who own both.
2. Step 2
  B1 Divide by the whole group of $25$ .
  $\dfrac{10 + 3}{25} = \dfrac{13}{25}$
Answer
$\dfrac{13}{25}$
Examiner tip
Examiner tip: A set in a Venn diagram is the whole circle, overlap included, not just its "only" region.

Common mistake: Giving $\frac{10}{25}$ , which is the probability of owning a scooter only.
6Two-way table — overall and given a row [2 marks]
Getting started• Paper 1H/2H style — 2 marks• two-way table, conditional probability
▼
Question
A gardener planted $120$ seeds, $60$ in compost X and $60$ in compost Y. The table shows how many germinated.

Germinated Did not germinate Total
Compost X $46$ $14$ $60$
Compost Y $33$ $27$ $60$
Total $79$ $41$ $120$

(a) One of the seeds is chosen at random. Find the probability that it did not germinate. [1]

(b) One of the seeds planted in compost Y is chosen at random. Find the probability that it germinated. [1]
Step-by-step solution
1. Step 1
  B1 (a) Any of the $120$ seeds can be chosen; the "did not germinate" column total is $41$ .
  $\dfrac{41}{120}$
2. Step 2
  B1 (b) Only the compost Y row is possible now, so the denominator is $60$ .
  $\dfrac{33}{60} = \dfrac{11}{20}$
Answer
(a) $\dfrac{41}{120}$ (b) $\dfrac{11}{20}$
Examiner tip
Examiner tip: Before dividing, underline the words that say which seeds could be picked. That group's total is the denominator.

Common mistake: Giving $\frac{33}{120}$ in (b), which is the probability that a seed was in compost Y and germinated.
7Probabilities in a Venn diagram, then given at least one [3 marks]
Building confidence• Paper 1H/2H style — 3 marks• venn diagram, set notation, conditional probability
▼
Question
$A$ and $B$ are two events. $P(A) = 0.7$ , $P(B) = 0.5$ and $P(A \cap B) = 0.35$

(a) Find $P(A' \cap B')$ . [2]

(b) Given that at least one of $A$ and $B$ occurs, find the probability that both occur. [1]
Step-by-step solution
1. Step 1
  M1 (a) The regions are $A$ only $0.35$ , both $0.35$ , $B$ only $0.15$ , so the probability of $A$ or $B$ or both is
  $P(A \cup B) = 0.7 + 0.5 - 0.35 = 0.85$
2. Step 2
  A1 $A' \cap B'$ is outside both circles.
  $P(A' \cap B') = 1 - 0.85 = 0.15$
3. Step 3
  B1 (b) "At least one" is $A \cup B$ , so $P(A \cup B) = 0.85$ from (a) is the denominator; "both" is the overlap.
  $\dfrac{P(A \cap B)}{P(A \cup B)} = \dfrac{0.35}{0.85} = \dfrac{7}{17}$
Answer
(a) $0.15$ (b) $\dfrac{7}{17}$
Examiner tip
Examiner tip: $A' \cap B'$ is "not $A$ and not $B$ ", the region outside both circles, which is the same set as $(A \cup B)'$ . In (b), reuse $P(A \cup B)$ rather than working it out again.

Common mistake: Working out $1 - 0.35 = 0.65$ in (a), which is $P((A \cap B)')$ . The dashes on both letters put the answer outside both circles, not just outside the overlap. In (b), leaving the answer as $0.35$ ignores the "given that".
8Two-way table from a ratio, then given that [3 marks]
Building confidence• Paper 1H/2H style — 3 marks• two-way table, ratio, conditional probability
▼
Question
A café sold $150$ drinks. Each drink was hot or cold, and small or large.

The ratio of hot drinks to cold drinks was $7 : 3$ .

$40$ of the hot drinks were large. $18$ of the cold drinks were small.

One of the large drinks is chosen at random. Find the probability that it was a cold drink. [3]
Step-by-step solution
1. Step 1
  M1 Share $150$ in the ratio $7 : 3$ .
  $150 \div 10 = 15, \qquad \text{hot} = 105, \quad \text{cold} = 45$
2. Step 2
  M1 Large cold drinks: $45 - 18 = 27$ , so large drinks altogether: $40 + 27 = 67$ .
3. Step 3
  A1 Given a large drink, the denominator is $67$ .
  $\dfrac{27}{67}$
Answer
$\dfrac{27}{67}$
Examiner tip
Examiner tip: Put the information in a two-way table (hot/cold by small/large) and fill in only what you need. Here the large column is enough.

Common mistake: Giving $\frac{27}{150}$ , the probability of a large cold drink out of all drinks. The drink is chosen from the large drinks only.
9Overlap from a given-that probability, then given exactly one [4 marks]
Stretch• Paper 1H/2H style — 4 marks• venn diagram, set notation, conditional probability
▼
Question
$A$ and $B$ are two events. $P(A) = 0.5$ and $P(B) = 0.35$

Given that $A$ occurs, the probability that $B$ also occurs is $0.4$ .

(a) Find $P(A' \cap B)$ . [2]

(b) Given that exactly one of the events $A$ and $B$ occurs, find the probability that it is $A$ . [2]
Step-by-step solution
1. Step 1
  M1 (a) "Given $A$ , the probability of $B$ is $0.4$ " means the overlap is $0.4$ of $P(A)$ .
  $P(A \cap B) = 0.4 \times 0.5 = 0.2$
2. Step 2
  A1 $A' \cap B$ is the $B$ only region.
  $P(A' \cap B) = 0.35 - 0.2 = 0.15$
3. Step 3
  M1 (b) $A$ only is $0.5 - 0.2 = 0.3$ . "Exactly one" is $A$ only together with $B$ only, $0.3 + 0.15 = 0.45$ , which is the given group.
4. Step 4
  A1 Of that group, $A$ only is the part where the event is $A$ .
  $\dfrac{0.3}{0.45} = \dfrac{2}{3}$
Answer
(a) $0.15$ (b) $\dfrac{2}{3}$
Examiner tip
Examiner tip: Put all four probabilities on a sketch Venn diagram ( $A$ only $0.3$ , both $0.2$ , $B$ only $0.15$ , neither $0.35$ ) and check they add up to $1$ before answering (b).

Common mistake: Taking $0.4$ as $P(A \cap B)$ itself. The $0.4$ is a fraction of the $A$ circle only, so it must be multiplied by $P(A) = 0.5$ ; using $0.4$ directly makes $B$ only $0.35 - 0.4 < 0$ , which is impossible.
10Two-way table from a percentage and a fraction [5 marks]
Stretch• Paper 1H/2H style — 5 marks• two-way table, percentages, conditional probability
▼
Question
$500$ adults in a town were asked whether they cycle to work. Each adult is under $30$ or aged $30$ and over.

$36\%$ of the adults cycle to work. $\frac{3}{5}$ of the adults who cycle to work are aged $30$ and over. $140$ of the adults are under $30$ .

(a) One of the adults is chosen at random. Find the probability that this adult is under $30$ and does not cycle to work. [3]

(b) One of the adults aged $30$ and over is chosen at random. Find the probability that this adult cycles to work. [2]
Step-by-step solution
1. Step 1
  M1 (a) Cyclists: $0.36 \times 500 = 180$ . Aged $30$ and over who cycle: $\frac{3}{5} \times 180 = 108$ , so under $30$ who cycle: $72$ .
2. Step 2
  M1 Under $30$ who do not cycle: $140 - 72 = 68$ .
3. Step 3
  A1 Divide by all $500$ adults.
  $\dfrac{68}{500} = \dfrac{17}{125}$
4. Step 4
  M1 (b) Aged $30$ and over: $500 - 140 = 360$ , of whom $108$ cycle.
5. Step 5
  A1 The denominator is the $360$ adults aged $30$ and over.
  $\dfrac{108}{360} = \dfrac{3}{10}$
Answer
(a) $\dfrac{17}{125}$ (b) $\dfrac{3}{10}$
Examiner tip
Examiner tip: The fraction $\frac{3}{5}$ is "of the adults who cycle", so it multiplies $180$ , not $500$ . Always ask: a fraction of which group?

Common mistake: Working out $\frac{3}{5} \times 500 = 300$ as the over-30 cyclists, which is more than the $180$ cyclists there are. A cell larger than its row or column total shows the wrong base was used.

	Germinated	Did not germinate	Total
Compost X	$46$	$14$	$60$
Compost Y	$33$	$27$	$60$
Total	$79$	$41$	$120$

Question 1
Paper 1H/2H style — 1 mark1 mark
A Venn diagram for sets $C$ and $D$ shows $12$ in $C$ only, $5$ in $C \cap D$ , $8$ in $D$ only and $15$ outside both sets.

One of the members of $\mathscr{E}$ is chosen at random. Find $P(C \cap D)$ . [1]
Model answer
$n(\mathscr{E}) = 12 + 5 + 8 + 15 = 40$

$P(C \cap D) = \frac{5}{40} = \frac{1}{8}$
Why this scores
Mark scheme:
- B1 for $\frac{5}{40}$ oe
Why this scores full marks: The total includes the $15$ outside both sets, and $C \cap D$ is only the overlap.
Question 2
Paper 1H/2H style — 2 marks2 marks
The table shows information about $200$ phone repairs.

Screen Battery Other Total
Under warranty $48$ $22$ $15$ $85$
Not under warranty $61$ $34$ $20$ $115$
Total $109$ $56$ $35$ $200$

One of the repairs is chosen at random. Find the probability that it was not a screen repair. [2]
Model answer
Not a screen repair $= 200 - 109 = 91$

$\frac{91}{200}$
Why this scores
Mark scheme:
- M1 for $200 - 109$ or $56 + 35$
- A1 for $\frac{91}{200}$ oe
Why this scores full marks: "Not a screen repair" is every column except Screen; using the complement of the Screen total is the quickest route.
Question 3
Paper 1H/2H style — 2 marks2 marks
$A$ and $B$ are two events. In a Venn diagram of probabilities, $P(A \cap B') = 0.28$ , $P(A \cap B) = 0.12$ and $P(A' \cap B) = 0.35$

Find $P(A' \cup B)$ . [2]
Model answer
$A' \cup B$ is every region except $A$ only.

$P(A' \cup B) = 1 - 0.28 = 0.72$
Why this scores
Mark scheme:
- M1 for identifying all regions except $A \cap B'$ , e.g. $1 - 0.28$ , or $0.12 + 0.35 + 0.25$
- A1 for $0.72$ oe
Why this scores full marks: The candidate reads $A' \cup B$ as "not $A$ , or $B$ " and sees that only the $A$ only region fails both conditions, so the unknown outside region never needs to be found.
Question 4
Paper 1H/2H style — 3 marks3 marks
There are $60$ people in a fitness class. $26$ of them are men. Each person is under $40$ or aged $40$ and over.

$20$ of the women are under $40$ . $22$ of the people are aged $40$ and over.

One of the people is chosen at random. Find the probability that the person is a man under $40$ . [3]
Model answer
Women $= 60 - 26 = 34$ , women aged $40$ and over $= 34 - 20 = 14$

Men aged $40$ and over $= 22 - 14 = 8$ , men under $40 = 26 - 8 = 18$

$\frac{18}{60} = \frac{3}{10}$
Why this scores
Mark scheme:
- M1 for $60 - 26 = 34$ and $34 - 20 = 14$
- M1 for $22 - 14 = 8$ and $26 - 8 = 18$ oe
- A1 for $\frac{18}{60}$ oe
Why this scores full marks: Each subtraction stays inside one row or one column of the two-way table, and the answer is checked: under $40$ totals $18 + 20 = 38 = 60 - 22$ .
Question 5
Paper 1H/2H style — 3 marks3 marks
$40$ people were asked whether they get news from TV ( $T$ ), radio ( $R$ ) or online ( $O$ ). The Venn diagram shows:
- $T$ only $10$ , $R$ only $8$ , $O$ only $6$
- $T$ and $R$ only $4$ , $T$ and $O$ only $3$ , $R$ and $O$ only $2$
- all three $1$ , none $6$
(a) A person is chosen at random. Find $P(O)$ . [1]

(b) Given that a person uses at least two of the sources, find the probability that the person uses all three. [2]
Model answer
(a) $n(O) = 6 + 3 + 2 + 1 = 12$

$P(O) = \frac{12}{40} = \frac{3}{10}$

(b) At least two $= 4 + 3 + 2 + 1 = 10$

$\frac{1}{10}$
Why this scores
Mark scheme:
- B1 (a) for $\frac{12}{40}$ oe
- M1 (b) for $4 + 3 + 2 + 1$ or $\frac{1}{k}$ with $k > 1$
- A1 (b) for $\frac{1}{10}$ oe
Why this scores full marks: "At least two" includes the centre as well as the three two-only regions, so the denominator is $10$ , not $9$ .
Question 6
Paper 1H/2H style — 3 marks3 marks
$A$ and $B$ are two events. $P(A) = 0.45$ , $P(B) = 0.3$ and $P(A \cap B) = 0.12$

Find the probability that $B$ occurs given that $A$ does not occur. [3]
Model answer
$P(A' \cap B) = 0.3 - 0.12 = 0.18$

$P(A') = 1 - 0.45 = 0.55$

$\frac{0.18}{0.55} = \frac{18}{55}$
Why this scores
Mark scheme:
- M1 for $0.3 - 0.12$ or $0.18$ in the $B$ only region
- M1 for $1 - 0.45$ or $0.55$ used as the denominator
- A1 for $\frac{18}{55}$ oe (e.g. awrt $0.327$ )
Why this scores full marks: "Given that $A$ does not occur" makes $P(A') = 0.55$ the denominator, and only the part of $B$ outside $A$ goes on top. A decimal answer of awrt $0.327$ is accepted, but the exact fraction $\frac{18}{55}$ is safest.
Question 7
Paper 1H/2H style — 4 marks4 marks
There are $60$ people in a Venn diagram for sets $A$ and $B$ . $9$ are in both sets and $12$ are in neither.

The number in $A$ only is twice the number in $B$ only.

One of the people not in $A$ is chosen at random. Find the probability that this person is in $B$ . [4]
Model answer
Let $B$ only $= y$ , so $A$ only $= 2y$ .

$2y + 9 + y + 12 = 60 \;\Rightarrow\; 3y = 39 \;\Rightarrow\; y = 13$

Not in $A$ : $B$ only $+$ neither $= 13 + 12 = 25$

$\frac{13}{25}$
Why this scores
Mark scheme:
- M1 for $2y + 9 + y + 12 = 60$ oe
- A1 for $y = 13$
- M1 for $\frac{k}{13 + 12}$ or $\frac{13}{m}$ with $m > 13$
- A1 for $\frac{13}{25}$ oe
Why this scores full marks: One letter for the smaller region keeps the equation simple, and the denominator is $A'$ , which is $B$ only together with the $12$ in neither.
Question 8
Paper 1H/2H style — 4 marks4 marks
$250$ people visited a zoo. $42\%$ of the visitors were children. $60\%$ of the children visited the aquarium. $95$ of the adults visited the aquarium.

Given that a visitor went to the aquarium, find the probability that the visitor was a child. [4]
Model answer
Children $= 0.42 \times 250 = 105$

Children at the aquarium $= 0.6 \times 105 = 63$

Aquarium visitors $= 63 + 95 = 158$

$\frac{63}{158}$
Why this scores
Mark scheme:
- M1 for $0.42 \times 250$ or $105$
- M1 for $0.6 \times 105$ or $63$
- M1 for $63 + 95$ or $158$
- A1 for $\frac{63}{158}$ oe
Why this scores full marks: The $60\%$ is applied to the children, not to all $250$ visitors, and the denominator is the whole aquarium group, children and adults together.
Question 9
Paper 1H/2H style — 5 marks5 marks
The Venn diagram for events $A$ and $B$ has these probabilities:

$P(A \cap B') = 3x$ , $P(A \cap B) = x$ , $P(A' \cap B) = 0.28$ , $P((A \cup B)') = 2x$

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

(b) Find the probability that $B$ occurs given that $A$ occurs. [2]

(c) Write down $P(A' \cap B')$ . [1]
Model answer
(a) $3x + x + 0.28 + 2x = 1 \;\Rightarrow\; 6x = 0.72 \;\Rightarrow\; x = 0.12$

(b) $P(A) = 3(0.12) + 0.12 = 0.48$

$\frac{0.12}{0.48} = \frac{1}{4}$

(c) $2x = 0.24$
Why this scores
Mark scheme:
- M1 (a) for $3x + x + 0.28 + 2x = 1$ oe
- A1 (a) for $0.12$
- M1 (b) for $\frac{x}{4x}$ or $\frac{0.12}{0.48}$ ft their $x$
- A1 (b) for $\frac{1}{4}$ oe
- B1 (c) for $0.24$ ft $2 \times$ their $x$
Why this scores full marks: The four probabilities sum to $1$ , giving the equation. In (b) the denominator is $P(A) = 0.48$ , the whole of circle $A$ , not $1$ .
Question 10
Paper 1H/2H style — 5 marks5 marks
A cinema sold $360$ tickets for three films: Action, Comedy and Drama. Each ticket was an adult ticket or a child ticket.

$\frac{5}{12}$ of the tickets were for Action, and $\frac{1}{3}$ of the Action tickets were child tickets.

$90$ tickets were for Comedy, and $54$ of these were child tickets.

$146$ of the tickets were child tickets.

(a) One of the child tickets is chosen at random. Find the probability that it was for Drama. [3]

(b) Given that a ticket was for Drama, find the probability that it was an adult ticket. [2]
Model answer
(a) Action $= \frac{5}{12} \times 360 = 150$ , Action child $= \frac{1}{3} \times 150 = 50$

Drama child $= 146 - 50 - 54 = 42$

$\frac{42}{146} = \frac{21}{73}$

(b) Drama $= 360 - 150 - 90 = 120$ , Drama adult $= 120 - 42 = 78$

$\frac{78}{120} = \frac{13}{20}$
Why this scores
Mark scheme:
- M1 (a) for $150$ and $50$
- M1 (a) for $146 - 50 - 54$ or $42$
- A1 (a) for $\frac{42}{146}$ oe
- M1 (b) for $360 - 150 - 90$ and $120 - 42$ , or $\frac{78}{k}$
- A1 (b) for $\frac{78}{120}$ oe
Why this scores full marks: The candidate keeps the two denominators apart: the child tickets ( $146$ ) in (a) and the Drama tickets ( $120$ ) in (b). Setting out a two-way table (films across, adult/child down) makes both totals visible.

4MA1 Venn Diagrams and Two-Way Tables

At a glance

What you’ll learn

Set notation

Venn diagrams (2 and 3 sets)

Probability from a Venn diagram

Two-way tables

Quick recap

Memorise this

How it’s examined

Worked ExamplesStep-by-step solutions to past-paper-style questions

1Fill a Venn diagram

2Probability from a Venn diagram

3Conditional probability from Venn

4Two-way table

5Probability of a set from Venn diagram counts [1 mark]

6Two-way table — overall and given a row [2 marks]

7Probabilities in a Venn diagram, then given at least one [3 marks]

8Two-way table from a ratio, then given that [3 marks]

9Overlap from a given-that probability, then given exactly one [4 marks]

10Two-way table from a percentage and a fraction [5 marks]

Model AnswersFull mark-scheme-style answers

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

	Screen	Battery	Other	Total
Under warranty	$48$	$22$	$15$	$85$
Not under warranty	$61$	$34$	$20$	$115$
Total	$109$	$56$	$35$	$200$