Compound-event sample spaces | Year 9

01 / LEARN

Understand the idea.

Record the whole experiment

A complete outcome records every stage of an experiment. For two coin tosses, HT means heads first, then tails. TH reverses that order, so these are different outcomes even though both contain exactly one head.

Build the sample space in a fixed order: start with H and list HH, HT; then start with T and list TH, TT. This gives every possible pair once. Counting only the final toss would hide half the experiment.

Check equal chances before counting

The counting rule $P (event) = \frac{favourable outcomes}{all outcomes}$ requires equally likely complete outcomes. For the fair coin and independent equal three-section spinner, the table has six such outcomes, each with probability $\frac{1}{6}$ .

An event can contain several complete outcomes. In two fair independent tosses, exactly one head contains HT and TH, whereas no heads contains only TT. The three head-count labels therefore do not have equal probabilities.

Fair coin followed by equal spinner 1, 2, 3
Coin / spinner	1	2	3
H	H1	H2	H3
T	T1	T2	T3

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Two fair coins

Find the probability of exactly one head in two fair independent tosses.

Worked solution

1
List the complete ordered outcomes.
$S = {HH, HT, TH, TT}$
Each toss has two possible results, giving four equally likely pairs for fair independent tosses.
2
Select the pairs with exactly one head.
$E = {HT, TH}$
HH has two heads and TT has none, so neither belongs to the event.
3
Divide the favourable count by the total count.
$P (exactly one head) = \frac{2}{4} = \frac{1}{2}$
Two of the four equally likely complete outcomes meet the condition.

Answer

$2 / 4 = 1 / 2$

EXAMPLE 02

Coin and spinner

For the displayed coin-spinner experiment, find the probability of a tail and an odd number.

Fair coin followed by equal spinner 1, 2, 3
Coin / spinner	1	2	3
H	H1	H2	H3
T	T1	T2	T3

Worked solution

1
Count the complete outcomes in the table.
$2 \times 3 = 6$
Each of the two coin results can occur with each of the three spinner results.
2
Find the outcomes satisfying both conditions.
$E = {T 1, T 3}$
The coin must be T, and the odd spinner labels are 1 and 3.
3
Calculate the probability.
$P (tail and odd) = \frac{2}{6} = \frac{1}{3}$
The fair independent devices make all six table entries equally likely.

Answer

$2 / 6 = 1 / 3$

03 / YOUR TURN

Try it yourself.

Check your understanding

What is P(no heads) in two fair independent tosses?

Show answer & reasoning

$1 / 4$

No heads requires TT. It is one outcome among the four equally likely pairs HH, HT, TH and TT, so $P (no heads) = \frac{1}{4}$ .

Keep practising

6 worksheet questions · 12 marks

Open worksheet Worked solutions

Read the questions online

List the sample space for two fair independent coin tosses.
1 mark
Find $P (two heads)$ for two fair independent coin tosses.
2 marks
A fair coin and an independent fair die are used. How many ordered outcomes are possible?
2 marks
For a fair coin and independent fair die, find $P (heads and die greater than 4)$ .
2 marks
An equal spinner labelled A, B, C is spun twice independently. Find $P (same letter)$ .
2 marks
Explain why zero, one and two heads are not equally likely in two fair independent tosses.
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

List the sample space for two fair independent coin tosses.
1 mark
Keep the first result H for two entries, then T.
HH, HT, TH, TT.
Find $P (two heads)$ for two fair independent coin tosses.
2 marks
Only HH is favourable out of four.
$\frac{1}{4}$
A fair coin and an independent fair die are used. How many ordered outcomes are possible?
2 marks
2 coin results times 6 die results gives 12.
$12$
For a fair coin and independent fair die, find $P (heads and die greater than 4)$ .
2 marks
H5 and H6 are favourable among 12 outcomes: $\frac{2}{12} = \frac{1}{6}$ .
$\frac{1}{6}$
An equal spinner labelled A, B, C is spun twice independently. Find $P (same letter)$ .
2 marks
AA, BB and CC are favourable among nine outcomes.
$\frac{1}{3}$
Explain why zero, one and two heads are not equally likely in two fair independent tosses.
3 marks
Group HH, HT, TH, TT by head count.
Their probabilities are $\frac{1}{4}$ , $\frac{1}{2}$ and $\frac{1}{4}$ .
Exactly one head has two ordered outcomes; zero and two heads have one each.

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Curriculum reference & lesson scope

Australian Curriculum v9 / SA Curriculum Mathematics

Year 9 compound-event sample spaces, probabilities, experiments and simulations.

Official curriculum source ↗

In this lesson

Understand the idea.

Record the whole experiment

Check equal chances before counting

See the method in action.

Two fair coins

List the complete ordered outcomes.

Select the pairs with exactly one head.

Divide the favourable count by the total count.

Coin and spinner

Count the complete outcomes in the table.

Find the outcomes satisfying both conditions.

Calculate the probability.

Try it yourself.

Check your understanding

Keep practising