Simulations of simple versus related compound events | Year 9

01 / LEARN

Understand the idea.

Make random digits match the model

A simulation must give outcomes the intended probabilities. For a fair coin, assign digits 0 to 4 to H and 5 to 9 to T. Independent equally likely digits then give equal chances of H and T; two digits form one two-toss trial.

For three equally likely spinner results, assign 0 to 2, 3 to 5 and 6 to 8 to the three outcomes. Reject 9 and read again. Each accepted group contains three digits, so each outcome has probability $\frac{3}{9} = \frac{1}{3}$ .

Reset between full trials, not between dependent stages

Without replacement, use individual labels R1, R2, R3, B1 and B2 so every counter has the same selection chance. Keep the first selected label unavailable for the second draw. Reset all five labels only after recording the complete two-draw trial.

Define success before reading random results, then count successful complete trials and divide by all complete trials. Check both the estimated probability and the method: extra trials reduce typical random variation but cannot repair an incorrect replacement rule.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Read supplied digits

Use 0–4 = H and 5–9 = T. Read pairs 07, 26, 44, 91, 58, 30. Estimate P(exactly one head).

Worked solution

1
Translate each digit pair using the given mapping.
$\begin{aligned} 07 & \to HT & 26 & \to HT \\ 44 & \to HH & 91 & \to TH \\ 58 & \to TT & 30 & \to HH \end{aligned}$
Digits 0 to 4 represent H, digits 5 to 9 represent T, and each supplied pair forms one trial.
2
Count the trials with exactly one head.
$HT, HT, TH : 3 successes$
The first, second and fourth trials have one H and one T; the repeated HT result still comes from two separate trials.
3
Calculate the relative-frequency estimate.
$\frac{3}{6} = \frac{1}{2}$
There are three successful complete trials out of six; this is an experimental estimate from the supplied digits.

Answer

$3 / 6 = 1 / 2$ .

EXAMPLE 02

A without-replacement simulation

Describe one fair simulated trial for two draws without replacement from R1, R2, R3, B1, B2.

Worked solution

1
Randomly order all five individual labels.
$R 1, R 2, R 3, B 1, B 2$
A shuffle in which every ordering is equally likely gives each counter an equal chance of appearing first.
2
Take the first two labels and record their colours.
$5 \times 4 = 20 possible ordered pairs$
The second label must differ from the first, modelling removal; every ordered pair has the same chance under the fair shuffle.
3
Keep both selections together as one complete trial.
Restoring the first counter before the second selection would change the experiment to sampling with replacement.
4
Restore all five labels before the next trial.
Each new trial must begin with the original three red and two blue counters so it repeats the same experiment.

Answer

Two distinct labels per trial, then reset all five labels between trials.

03 / YOUR TURN

Try it yourself.

Check your understanding

In the equal-three-outcome digit mapping, what happens to digit 9?

Show answer & reasoning

Reject it and draw again.

Reject digit 9 and read a fresh digit. The accepted digits 0 to 8 split into three groups of equal size, giving each outcome probability $\frac{1}{3}$ ; assigning 9 to one group would make that outcome more likely.

Keep practising

6 worksheet questions · 13 marks

Open worksheet Worked solutions

Read the questions online

Why is $0 – 5 = H$ and $6 – 9 = T$ not a fair-coin model?
2 marks
Map 0–4 to H and 5–9 to T. Convert 82, 16, 73, 40 into outcomes.
2 marks
The simulated outcomes are TH, HT, TH, HH. Estimate $P (exactly one head)$ .
2 marks
Give a random-digit model for success probability 0.3.
2 marks
A without-replacement simulation selects the same labelled counter twice in a trial. What needs fixing?
2 marks
An event occurs 198 times in 500 correctly modelled trials. Give an estimate and one limitation.
3 marks

Read the worked solutions

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

Why is $0 – 5 = H$ and $6 – 9 = T$ not a fair-coin model?
2 marks
The groups have unequal sizes.
H has chance $\frac{6}{10}$ and T $\frac{4}{10}$ .
Map 0–4 to H and 5–9 to T. Convert 82, 16, 73, 40 into outcomes.
2 marks
Classify each digit, keeping its order.
TH, HT, TH, HH.
The simulated outcomes are TH, HT, TH, HH. Estimate $P (exactly one head)$ .
2 marks
TH, HT and TH are successes.
$\frac{3}{4}$
Give a random-digit model for success probability 0.3.
2 marks
Three of ten equally likely digits mean success.
For example, $0 – 2 = success$ and $3 – 9 = failure$ .
A without-replacement simulation selects the same labelled counter twice in a trial. What needs fixing?
2 marks
A physical counter cannot be selected while it is kept out.
Remove the first selected label before selecting the second.
An event occurs 198 times in 500 correctly modelled trials. Give an estimate and one limitation.
3 marks
$\frac{198}{500} = 0.396$ .
More trials help stability, but the model must also fit the real rules.
$0.396$ ; a finite simulation still varies randomly.

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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.

Make random digits match the model

Reset between full trials, not between dependent stages

See the method in action.

Read supplied digits

Translate each digit pair using the given mapping.

Count the trials with exactly one head.

Calculate the relative-frequency estimate.

A without-replacement simulation

Randomly order all five individual labels.

Take the first two labels and record their colours.

Keep both selections together as one complete trial.

Restore all five labels before the next trial.

Try it yourself.

Check your understanding

Keep practising