Sample spaces, complementary events, arrays and tree diagrams for two-stage events (Year 12 Relative frequency and probability, 2024 Mathematics Standard 1 syllabus)

Syllabus dot point

“Identify the sample space of an experiment, find the probability of an event with equally likely outcomes and its complement, and use arrays and tree diagrams to solve problems about two-stage events”

9 min readUpdated 7 Oct 2026

Quick answer

List the sample space (an array for two dice), then P(event) = favourable over total, and P(not event) = 1 minus P(event). For two-stage events, draw a tree diagram, multiply along each path and add the paths that give the event; with replacement the second stage repeats the first. For "at least one", use 1 minus P(none). New as its own Year 12 focus area in the 2024 Mathematics Standard 1 syllabus.

Jump to a section

What this dot point is asking
The answer
Practice questions

What this dot point is asking

In the Mathematics Standard 11-12 Syllabus (2024), first examined in the 2027 HSC, Year 12 Relative frequency and probability is a focus area in its own right. Probability used to be Year 11 work; under the new syllabus it is part of Year 12 Standard 1. NESA's Year 12 Standard 1 page asks you to identify the sample space of an experiment, express the probability of an event with equally likely outcomes as favourable over total, use the complement, and use arrays and tree diagrams to solve problems about two-stage events, involving replacement where applicable.

An array is a grid listing every outcome of the two stages, so a two-way grid of outcomes (one die down the side, the other across the top) is an array.

Two-stage means an experiment done in two steps: two dice, a coin then a die, or two counters drawn from a bag.

The answer

Key fact

For equally likely outcomes,

P(\text{event}) = \frac{\text{number of favourable outcomes}}{\text{total number of outcomes}}, \qquad 0 \le P \le 1.

The complement rule is $P(\text{event does not occur}) = 1 - P(\text{event occurs})$ .

On a tree diagram, multiply along the branches for one path, then add the paths that give the event you want.

Sample space

The sample space is the full list of outcomes. For one die it is $\{1, 2, 3, 4, 5, 6\}$ . For two dice there are $6 \times 6 = 36$ outcomes, which is far too many to list as pairs in a line, so use an array (a two-way grid) with one die down the side and the other across the top. Each cell is one outcome, and counting cells answers the question.

Tree diagrams

Draw one set of branches for the first stage, then a second set from the end of each first-stage branch. Label each branch with its probability.

A path from left to right is one complete outcome, such as RB (red then blue).
Replacement means the item is put back, so the second set of branches has the same probabilities as the first. This is the case the syllabus names.

For a bag with $4$ red and $6$ blue counters drawn with replacement, $P(R) = \frac{2}{5}$ and $P(B) = \frac{3}{5}$ on both draws, so $P(RR) = \frac{2}{5} \times \frac{2}{5} = \frac{4}{25}$ .

The complement shortcut

"At least one" questions are almost always quicker through the complement. At least one blue is everything except "no blue":

P(\text{at least one blue}) = 1 - P(\text{no blue}) = 1 - \frac{4}{25} = \frac{21}{25}.

Choosing a tool

Array: best when both stages have equally likely outcomes and you want to count favourable cells (two dice, two spinners).
Tree diagram: best when the branches have different probabilities (a bag with unequal colours) or when you want to multiply along paths.

Worked examples

Find a probability from a sample space table

A spinner has sectors $1, 2, 3, 4$ (equally likely) and is spun twice. To find $P(\text{sum} = 5)$ , the table has $16$ cells. The pairs adding to $5$ are $(1,4), (2,3), (3,2), (4,1)$ , so $P = \frac{4}{16} = \frac{1}{4}$ .

Marker's note: say how many cells are in the table, then count the favourable ones.

Multiply, then add

A bag has $3$ yellow and $7$ other counters. With replacement, $P(\text{exactly one yellow in two draws}) = \frac{3}{10} \times \frac{7}{10} + \frac{7}{10} \times \frac{3}{10} = \frac{21}{100} + \frac{21}{100} = \frac{42}{100} = \frac{21}{50}$ .

Marker's note: there are two paths (yellow first, or yellow second), so both are added.

Use the complement

$P(\text{at least one six in two rolls of a die}) = 1 - \frac{5}{6} \times \frac{5}{6} = 1 - \frac{25}{36} = \frac{11}{36}$ .

Marker's note: finding "no six on either roll" first is the efficient method.

Common traps

Adding along the branches: Multiply along a path, add between paths.
Missing a path: "One of each colour" has two paths (RB and BR), not one.
Treating outcomes as equally likely when they are not: A bag with $3$ yellow, $2$ green and $5$ blue does not give $\frac{1}{3}$ for each colour. Use the counts.
Giving a probability above $1$: A probability must be between $0$ and $1$ . If you get more than $1$ , you have added instead of multiplied or double counted.
Using the wrong denominator: The denominator is the size of the whole sample space (or the number in the bag), not the number of favourable outcomes.

Exam technique

Write the sample space size first ( $36$ for two dice, $10$ for a bag of $10$ ). Draw the tree even when you could do the sum in your head, because marks are given for the setup. For "at least one" always check whether $1 - P(\text{none})$ is shorter. Finish by checking that your outcome probabilities add to $1$ .

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 fair six-sided die is rolled once. Find the probability that the number rolled is a factor of

12

, and the probability that it is not a factor of

12

Show worked solution →

List the favourable outcomes. The sample space is $\{1, 2, 3, 4, 5, 6\}$ . The factors of $12$ in it are $1, 2, 3, 4, 6$ , which is $5$ outcomes.

P(\text{factor of } 12) = \frac{5}{6}.

Use the complement. $P(\text{not a factor}) = 1 - \frac{5}{6} = \frac{1}{6}$ (only $5$ is left).

Marker's note: one mark for $\frac{5}{6}$ , one for $\frac{1}{6}$ . Leaving out $1$ as a factor is the usual slip.

foundation3 marks

Two fair dice are rolled and the numbers are added. Draw an array (a two-way grid) of the sample space and use it to find the probability that the sum is

7

, and the probability that the sum is not

7

Show worked solution →

Build the table. Rows are the first die and columns the second die, and each cell holds the sum.

	1	2	3	4	5	6
1	2	3	4	5	6	7
2	3	4	5	6	7	8
3	4	5	6	7	8	9
4	5	6	7	8	9	10
5	6	7	8	9	10	11
6	7	8	9	10	11	12

There are $36$ equally likely outcomes, and a sum of $7$ appears $6$ times (the diagonal).

P(\text{sum } 7) = \frac{6}{36} = \frac{1}{6}, \qquad P(\text{not } 7) = 1 - \frac{1}{6} = \frac{5}{6}.

Marker's note: one mark for a correct table, one for $\frac{1}{6}$ , one for the complement.

core3 marks

A bag holds

4

red and

6

blue counters. A counter is drawn, its colour is noted and it is put back. A second counter is then drawn. Draw a tree diagram and find the probability that both counters are red.

Show worked solution →

Tree diagram. Because the first counter is replaced, both draws have the same probabilities: $P(R) = \frac{4}{10} = \frac{2}{5}$ and $P(B) = \frac{6}{10} = \frac{3}{5}$ .

First draw	Second draw	Outcome	Probability
R ( $\frac{2}{5}$ )	R ( $\frac{2}{5}$ )	RR	$\frac{4}{25}$
R ( $\frac{2}{5}$ )	B ( $\frac{3}{5}$ )	RB	$\frac{6}{25}$
B ( $\frac{3}{5}$ )	R ( $\frac{2}{5}$ )	BR	$\frac{6}{25}$
B ( $\frac{3}{5}$ )	B ( $\frac{3}{5}$ )	BB	$\frac{9}{25}$

Both red. Multiply along the branches: $P(RR) = \frac{2}{5} \times \frac{2}{5} = \frac{4}{25}$ .

Marker's note: one mark for the correct branch probabilities, one for the tree, one for $\frac{4}{25}$ . The four outcome probabilities add to $1$ , which is a useful check.

core3 marks

Using the same bag and the same replacement rule as the previous question, find the probability that (a) one counter of each colour is drawn, in either order, and (b) at least one counter is blue.

Show worked solution →

(a) One of each. The outcomes RB and BR both qualify, so add their branch products:

P(\text{one of each}) = \frac{6}{25} + \frac{6}{25} = \frac{12}{25}.

(b) At least one blue. The only outcome with no blue is RR, so use the complement:

P(\text{at least one blue}) = 1 - \frac{4}{25} = \frac{21}{25}.

Check by adding RB, BR and BB: $\frac{6 + 6 + 9}{25} = \frac{21}{25}$ .

Marker's note: one mark for adding both orders in (a), one for each of the two parts' final values. Forgetting the second order is the usual error in (a).

core3 marks

A coin is tossed and then a fair die is rolled. Find the probability of a head followed by an even number, and the probability of a tail followed by a number greater than

4

Show worked solution →

The two stages do not affect each other, so multiply.

P(\text{head and even}) = \frac{1}{2} \times \frac{3}{6} = \frac{1}{4}.

The even numbers are $2, 4, 6$ , which is $3$ out of $6$ .

P(\text{tail and} > 4) = \frac{1}{2} \times \frac{2}{6} = \frac{1}{6}.

The numbers greater than $4$ are $5$ and $6$ , which is $2$ out of $6$ .

Marker's note: one mark for each correct product and one for identifying the favourable outcomes correctly.

exam4 marks

The probability of rain on Monday is

0.3

and the probability of rain on Tuesday is

0.2

. Assume that the weather on the two days is independent. Find the probability that (a) it rains on both days, (b) it rains on at least one of the two days, and (c) it rains on exactly one of the two days.

Show worked solution →

Set up: $P(R_M) = 0.3$ , so $P(\text{dry Monday}) = 0.7$ . $P(R_T) = 0.2$ , so $P(\text{dry Tuesday}) = 0.8$ .
(a) Both days: $0.3 \times 0.2 = 0.06$ .
(b) At least one day: The complement is dry on both days: $0.7 \times 0.8 = 0.56$ . So

P(\text{at least one}) = 1 - 0.56 = 0.44.

(c) Exactly one day. Rain Monday and dry Tuesday, or dry Monday and rain Tuesday:

0.3 \times 0.8 + 0.7 \times 0.2 = 0.24 + 0.14 = 0.38.

Check: $0.06 + 0.38 = 0.44$ .

Marker's note: one mark per part, and one for stating the independence assumption. The check in (c) shows (a), (b) and (c) are consistent.

exam4 marks

A bag contains

3

yellow,

2

green and

5

blue counters. A counter is drawn and replaced, then a second counter is drawn. Find the probability that (a) both counters are the same colour, (b) the counters are different colours, and (c) neither counter is green.

Show worked solution →

Single-draw probabilities. $P(Y) = \frac{3}{10}$ , $P(G) = \frac{2}{10}$ , $P(B) = \frac{5}{10}$ .

(a) Same colour. Add the three matching outcomes:

\frac{3}{10} \times \frac{3}{10} + \frac{2}{10} \times \frac{2}{10} + \frac{5}{10} \times \frac{5}{10} = \frac{9 + 4 + 25}{100} = \frac{38}{100} = \frac{19}{50}.

(b) Different colours. This is the complement of (a): $1 - \frac{19}{50} = \frac{31}{50}$ .

(c) Neither green. Each draw is not green with probability $\frac{8}{10}$ , so

\frac{8}{10} \times \frac{8}{10} = \frac{64}{100} = \frac{16}{25}.

Marker's note: one mark for each part, and one for the correct single-draw probabilities. Using the complement in (b) saves listing six outcomes.

Practise this

Sources & how we know this

Mathematics Standard 11-12 Syllabus (2024): Year 12 Standard 1 Relative frequency and probability — NESA (2024)
Mathematics Standard 11-12 Syllabus (2024): course overview and implementation — NESA (2024)
Mathematics Standard Stage 6 Syllabus (2017) — NESA
Mathematics Standard HSC exam papers — NESA