Designing and interpreting simulations | Year 10

01 / LEARN

Understand the idea.

Make the simulation match the experiment

A simulation must reproduce the experiment's relevant probabilities and restrictions. Uniform random digits can model a sixty per cent success chance by assigning six digits, zero through five, to success and the remaining four to failure.

Model each physical stage separately when its probabilities depend on that stage. For fair dice, generate two independent values from one to six and add them. For sampling without replacement, remove selected items inside a trial and restore the original collection between trials.

Interpret estimates and variability

Estimate probability using relative frequency: divide the event count by the number of trials. Define one complete trial and the success event in advance so the numerator and denominator describe the same experiment.

Finite runs fluctuate. More trials generally produce more stable estimates but do not guarantee an exact answer or improvement in every run. Report the estimate, trial count and model assumptions; simulation cannot correct a model that gives the wrong underlying outcomes.

relative frequency = \frac{event count}{trial count}

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

A digit simulation

Use 0–5 for success in digits 0,7,4,6,3,1,9,5,2,8. Find the relative frequency.

Worked solution

1
Apply the stated success rule to each digit.
$0, 4, 3, 1, 5, 2$
These six recorded digits lie between zero and five inclusive.
2
Count successes and trials.
$N_{success} = 6, N_{trials} = 10$
Every recorded digit represents one trial, including failures.
3
Calculate the observed relative frequency.
$\hat{p} = \frac{6}{10} = 0.6$
This run has a sixty per cent success rate; matching the model probability exactly here does not guarantee future runs will.

Answer

6/10=0.6.

EXAMPLE 02

A flawed dice simulation

A student chooses sums 2–12 uniformly. Why is this wrong for two fair dice?

Worked solution

1
Identify the probability assigned by the proposed generator.
$P_{generator} (each sum) = \frac{1}{11}$
There are eleven integers from two to twelve, and the generator chooses them uniformly.
2
Compare actual dice outcome counts.
$P (sum 2) = \frac{1}{36}, P (sum 7) = \frac{6}{36}$
Only $(1, 1)$ gives two, whereas six distinct ordered pairs give seven.
3
Replace the generator with the correct experiment.
$D_{1}, D_{2} \in {1, 2, 3, 4, 5, 6}, S = D_{1} + D_{2}$
Generate the two fair die results independently, then add them so sum probabilities follow their actual pair counts.

Answer

Generate two independent die results instead.

03 / YOUR TURN

Try it yourself.

Check your understanding

A simulation produces 126 successes in 200 trials. Estimate the probability.

Show answer & reasoning

0.63.

The relative frequency is $\hat{p} = \frac{126}{200} = 0.63$ . Report it as the estimated success probability from these two hundred trials.

Keep practising

6 worksheet questions · 15 marks

Open worksheet Worked solutions

Read the questions online

Describe a random-digit simulation of a 30% chance.
2 marks
A simulation gives 47 events in 250 trials. Find relative frequency.
2 marks
Explain why generating one uniform integer 2–12 does not simulate two dice sums.
2 marks
Design a simulation of two counters drawn without replacement from 3 red and 2 blue.
3 marks
Runs of 100 and 1000 trials estimate $P (A)$ as 0.66 and 0.604. If the theoretical probability is 0.6, which is closer and by how much?
3 marks
Does a larger run guarantee that every 1000-trial simulation estimate is closer to the theoretical probability than every 100-trial estimate? Explain.
3 marks

Read the worked solutions

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

Describe a random-digit simulation of a 30% chance.
2 marks
Three of ten uniform digits represent success.
Use 0,1,2 for success and 3–9 for failure.
A simulation gives 47 events in 250 trials. Find relative frequency.
2 marks
$\frac{47}{250} = 0.188$ .
0.188.
Explain why generating one uniform integer 2–12 does not simulate two dice sums.
2 marks
Sum 7 has 6 pairs;sum 2 has 1.
Real sums have unequal numbers of ordered pairs.
Design a simulation of two counters drawn without replacement from 3 red and 2 blue.
3 marks
This prevents reuse within a trial.
Resetting between trials restores the original experiment.
Shuffle five labelled items, three red and two blue, then take the first two; reshuffle for each new trial.
Runs of 100 and 1000 trials estimate $P (A)$ as 0.66 and 0.604. If the theoretical probability is 0.6, which is closer and by how much?
3 marks
Calculate $∣ 0.66 - 0.6 ∣ = 0.06$ and $∣ 0.604 - 0.6 ∣ = 0.004$ .
Compare their errors: $0.06 - 0.004 = 0.056$ .
The 1000-trial estimate is closer. The absolute errors are 0.06 and 0.004, so its error is smaller by 0.056.
Does a larger run guarantee that every 1000-trial simulation estimate is closer to the theoretical probability than every 100-trial estimate? Explain.
3 marks
It compares only two particular runs.
Random fluctuations can make a larger finite run less accurate.
No.

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Curriculum reference & lesson scope

Australian Curriculum v 9 / SA Curriculum

Compound events; systematic counting and the multiplication principle; factorials and simple arrangements; conditional language and restricted sample spaces; conditional probabilities from tables/trees; replacement and dependence; conditional versus unconditional probabilities; designing and interpreting simulations.

Official curriculum source ↗

In this lesson

Understand the idea.

Make the simulation match the experiment

Interpret estimates and variability

See the method in action.

A digit simulation

Apply the stated success rule to each digit.

Count successes and trials.

Calculate the observed relative frequency.

A flawed dice simulation

Identify the probability assigned by the proposed generator.

Compare actual dice outcome counts.

Replace the generator with the correct experiment.

Try it yourself.

Check your understanding

Keep practising