Venn diagrams and two-way tables for two attributes: filling in the regions, converting between the two displays, and finding probabilities and proportions (new in the 2024 Mathematics Standard syllabus)

Syllabus dot point

“Construct and interpret Venn diagrams and two-way tables from given information, limited to combinations of two attributes, and use them to solve problems in a variety of contexts”

HSCMaths Standard 2Year 12: Relative Frequency and…11 min readUpdated 23 Sep 2026

Quick answer

For two attributes, a Venn diagram has four regions (A only, both, B only, neither) and a two-way table has the same four cells. Fill "both" first, subtract it from each total, then find neither from the grand total. Probabilities are region counts over the total, except "of those who ..." questions, which divide by that group. Part of Year 12 Relative frequency and probability in the 2024 Mathematics Standard syllabus.

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What this dot point is asking
The answer
Exam-style questions
Practice questions

What this dot point is asking

In the Mathematics Standard 11-12 Syllabus (2024), first examined in the 2027 HSC, relative frequency and probability moved from Year 11 to Year 12, and it now includes this content point: "construct and interpret Venn diagrams and two-way tables from given information, limited to combinations of two attributes, and use them to solve problems in a variety of contexts". Two-way tables appeared in the 2017 course (for example in expected-frequency problems), but Venn diagrams are new to Mathematics Standard.

"Two attributes" means every question involves exactly two yes/no characteristics (plays sport or not, owns a dog or not), so every diagram has two circles and every table has two rows and two columns of data.

Note

Think of a class photo where everyone who plays sport stands in one hoop on the floor and everyone who plays music stands in another. The hoops overlap, and the people who do both stand in the overlap. Anyone who does neither stands outside both hoops. Counting the people in each part of the floor tells you everything: how many do sport, how many do music, how many do both and how many do neither. A two-way table is the same count written in a grid.

The answer

Key fact

A Venn diagram for two attributes $A$ and $B$ has four regions: $A$ only, both, $B$ only, and neither. Fill the overlap first, then subtract it from each circle total:

A \text{ only} = n(A) - \text{both}, \qquad n(A \text{ or } B) = n(A) + n(B) - \text{both}, \qquad \text{neither} = \text{total} - n(A \text{ or } B).

Probabilities are counts over the relevant total. For "of those who have $A$ ", divide by $n(A)$ , not by the whole total.

Filling in a Venn diagram

Take a class of $120$ students where $70$ play sport, $45$ play music and $25$ do both.

Both: $25$ .
Sport only: $70 - 25 = 45$ . Music only: $45 - 25 = 20$ .
At least one: $45 + 25 + 20 = 90$ . Neither: $120 - 90 = 30$ .

Now read off anything the question asks: $P(\text{sport or music}) = \frac{90}{120} = \frac{3}{4}$ , $P(\text{neither}) = \frac{30}{120} = \frac{1}{4}$ , $P(\text{music only}) = \frac{20}{120} = \frac{1}{6}$ , and of the $45$ music students, $\frac{25}{45} = \frac{5}{9}$ also play sport.

The same information as a two-way table

	Music	No music	Total
Sport	$25$	$45$	$70$
No sport	$20$	$30$	$50$
Total	$45$	$75$	$120$

Each inner cell is one Venn region: "sport and music" is the overlap, "sport, no music" is sport only, "no sport, music" is music only, and "no sport, no music" is neither. Rows and columns must add to their totals, which is a built-in check.

Choosing the right denominator

Most lost marks come from dividing by the wrong total:

"What is the probability that a student plays music?" uses the whole group: $\frac{45}{120}$ .
"Of the students who play music, what proportion play sport?" uses only the music students: $\frac{25}{45}$ .

The second kind of question is how statistics are used (and misused) in the media, which the syllabus also asks you to examine.

How exam questions ask about it

"Complete the Venn diagram / two-way table." Start from "both", then the "only" regions, then neither.
"How many have neither?" or "how many have both?" Use $n(A \text{ or } B) = n(A) + n(B) - \text{both}$ and the total.
"Find the probability that ..." Count the region(s) and divide by the total, unless the question restricts the group.
"Of those who ..., what proportion ...?" Divide by that group's total.
"Estimate the number in a population." Multiply the relative frequency by the population size (expected frequency $np$ ).

Worked examples

Find a missing overlap

In a class of $60$ , $35$ study Legal Studies, $28$ study Business Studies and $9$ study neither. At least one: $60 - 9 = 51$ . The circle totals add to $63$ , which counts the overlap twice, so both $= 63 - 51 = 12$ . Then Legal only $= 23$ , Business only $= 16$ .

Marker's note: showing " $63 - 51 = 12$ " with a reason (double counting) earns the method mark.

Build a two-way table and read a proportion

$300$ vaccinated adults include $15$ who caught the flu; $100$ unvaccinated adults include $20$ who caught it. The table has rows vaccinated / not vaccinated and columns flu / no flu: $15, 285$ and $20, 80$ , with column totals $35$ and $365$ . Of the $35$ who caught the flu, $\frac{15}{35} = \frac{3}{7}$ were vaccinated, but the rates within each group are $5\%$ and $20\%$ .

Marker's note: when comparing two groups of different sizes, compare rates within each group, not raw counts.

From percentages

If $60\%$ exercise, $30\%$ smoke and $10\%$ do both, then exercise only $50\%$ , smoke only $20\%$ , neither $20\%$ . For $500$ adults that is $100$ adults in neither.

Marker's note: percentages fill a Venn diagram exactly like counts; convert to numbers at the end.

Common traps

Putting the circle total in the "only" region: If $70$ play sport and $25$ do both, "sport only" is $45$ , not $70$ .
Adding circle totals to get "at least one": $70 + 45 = 115$ double counts the $25$ in both; the correct count is $90$ .
Forgetting the "neither" region: It must be included for the diagram to add to the total.
Wrong denominator: "Of those who ..." restricts the group; divide by that group's size.
Reading "or" as "only one": In probability " $A$ or $B$ " includes both.

Exam technique

Draw the diagram or table even if the question does not ask for it, and write the overlap first. Check that the four regions add to the total before answering. Circle the words "of those who" in a question: they tell you to change the denominator. Give probabilities as simplified fractions or decimals, and multiply by the population size for "estimate how many" questions.

Exam-style questions

Questions in the style of NESA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

HSC-style3 marks

In a group of

90

students,

48

study Biology,

36

study Chemistry and

18

study both. (a) How many students study neither subject? (b) A student is chosen at random. Find the probability that they study exactly one of the two subjects.

Show worked answer →

Biology only $= 48 - 18 = 30$ and Chemistry only $= 36 - 18 = 18$ , so $30 + 18 + 18 = 66$ study at least one.

(a) Neither $= 90 - 66 = 24$ .

(b) Exactly one $= 30 + 18 = 48$ , so $P = \frac{48}{90} = \frac{8}{15}$ .

Markers look for the overlap subtracted from each circle total, a correct count of neither, and the probability using the whole group of $90$ as the denominator.

HSC-style3 marks

A survey of

250

phone users recorded the type of phone and whether the user was under

25

. Of the

140

Android users,

60

were under

25

. Of the

110

iPhone users,

70

were under

25

. (a) Find the probability that a randomly chosen user is under

25

. (b) Of the users under

25

, what proportion use an iPhone?

Show worked answer →

Under $25$ : $60 + 70 = 130$ .

(a) $P(\text{under } 25) = \frac{130}{250} = \frac{13}{25} = 0.52$ .

(b) Restrict to the $130$ users under $25$ : $\frac{70}{130} = \frac{7}{13} \approx 0.54$ .

Markers expect a two-way table (or equivalent working) and, in (b), the denominator $130$ , not $250$ .

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation2 marks

A Venn diagram for two attributes

A

and

B

shows

12

in "

A

only",

8

in "both",

15

in "

B

only" and

5

outside both circles. Find the total number of people and the probability that a randomly chosen person has attribute

A

Show worked solution →

Total. Add all four regions: $12 + 8 + 15 + 5 = 40$ .

Has $A$ . The $A$ circle is " $A$ only" plus "both": $12 + 8 = 20$ , so

P(A) = \frac{20}{40} = \frac{1}{2}.

Marker's note: one mark for $40$ , one for $\frac{1}{2}$ . Using $12$ alone (forgetting the overlap) is the usual error.

foundation2 marks

50

people,

30

like tea,

25

like coffee and

10

like both. Complete a Venn diagram and state how many like neither.

Show worked solution →

Start with the overlap. Both: $10$ . Tea only: $30 - 10 = 20$ . Coffee only: $25 - 10 = 15$ .

Neither. $50 - (20 + 10 + 15) = 5$ .

Marker's note: one mark for the three inner regions, one for $5$ .

core3 marks

In a survey of

200

drivers,

80

had received a speeding fine,

50

a parking fine and

30

both. Show this information in a two-way table, and find the probability that a randomly chosen driver had received neither kind of fine.

Show worked solution →

Fill the table from the overlap.

	Parking fine	No parking fine	Total
Speeding fine	$30$	$50$	$80$
No speeding fine	$20$	$100$	$120$
Total	$50$	$150$	$200$

(Speeding only $= 80 - 30 = 50$ ; parking only $= 50 - 30 = 20$ ; neither $= 200 - 30 - 50 - 20 = 100$ .)

Probability. $P(\text{neither}) = \frac{100}{200} = \frac{1}{2}$ .

Marker's note: one mark for the "both" cell and the two "only" cells, one for a consistent table with totals, one for $\frac{1}{2}$ .

core3 marks

In a class of

60

students,

35

study Legal Studies,

28

study Business Studies and

9

study neither. How many study both? Find the probability that a randomly chosen student studies exactly one of the two subjects.

Show worked solution →

At least one: $60 - 9 = 51$ students study at least one subject.
Both: The two circle totals add to $35 + 28 = 63$ , which counts the overlap twice, so both $= 63 - 51 = 12$ .
Exactly one: Legal only $= 35 - 12 = 23$ and Business only $= 28 - 12 = 16$ , so $23 + 16 = 39$ students and

P(\text{exactly one}) = \frac{39}{60} = \frac{13}{20}.

Marker's note: one mark for $12$ (reasoning about the double count), one for the two "only" regions, one for $\frac{13}{20}$ .

core3 marks

A health survey of

400

adults recorded whether each was vaccinated and whether they caught the flu.

300

were vaccinated, and

15

of them caught the flu. Of the

100

unvaccinated adults,

20

caught the flu. (a) Of the adults who caught the flu, what proportion were vaccinated? (b) Compare the rate of catching the flu in the two groups.

Show worked solution →

(a) Restrict to the flu row. $15 + 20 = 35$ caught the flu, and $15$ of these were vaccinated: $\frac{15}{35} = \frac{3}{7}$ .

(b) Compare within each group. Vaccinated: $\frac{15}{300} = 5\%$ . Unvaccinated: $\frac{20}{100} = 20\%$ . The unvaccinated group caught the flu at four times the rate.

Marker's note: one mark for (a); two marks for (b), which needs both rates. The trap is to compare $15$ with $20$ directly, ignoring that the groups are different sizes.

exam4 marks

A survey of

400

households in a town found that

140

own a dog,

90

own a cat and

40

own both. (a) Find the probability that a household owns a dog or a cat. (b) The town has

12\,000

households. Estimate how many own neither a dog nor a cat.

Show worked solution →

(a) At least one. Dog or cat $= 140 + 90 - 40 = 190$ , so the relative frequency is

\frac{190}{400} = \frac{19}{40} = 0.475.

(b) Neither in the sample. $400 - 190 = 210$ , a relative frequency of $\frac{210}{400} = 0.525$ .

Expected number. $12\,000 \times 0.525 = 6300$ households.

Marker's note: one mark for $190$ , one for $0.475$ , one for $0.525$ , one for $6300$ (expected frequency $= np$ ).

exam4 marks

A report on

500

adults states: "60% exercise regularly, 30% smoke, and 10% both exercise regularly and smoke." (a) How many of the adults neither exercise regularly nor smoke? (b) A headline says "Most smokers exercise regularly." Use the data to decide whether this is supported.

Show worked solution →

(a) Percentages to a Venn diagram. Both: $10\%$ . Exercise only: $60\% - 10\% = 50\%$ . Smoke only: $30\% - 10\% = 20\%$ . Neither: $100\% - 80\% = 20\%$ , which is $0.2 \times 500 = 100$ adults.

(b) Look only at the smokers. Smokers are $30\%$ of the adults, and $10\%$ of the adults are smokers who exercise. So the proportion of smokers who exercise is $\frac{10}{30} = \frac{1}{3}$ . Only a third of smokers exercise regularly, so the headline is not supported.

Marker's note: two marks for (a); two for (b), which must restrict attention to the smokers (the $30\%$ ), not all $500$ adults.

Practise this

Sources & how we know this

Mathematics Standard 11-12 Syllabus (2024): Year 12 Standard 2 Relative frequency and probability — NESA (2024)
Mathematics Standard Stage 6 Syllabus (2017) — NESA
Mathematics Standard HSC exam papers — NESA