The idea, step by step
Make random digits match the model
For a fair coin, digits 0–4 can mean H and 5–9 mean T, using independent equally likely digits. Read two digits per trial to simulate two tosses. Decide the target event before reading the digits.
For an equal three-section spinner, map 0–2, 3–5 and 6–8 to its outcomes. Reject 9 and read another digit. Assigning 6–9 to the final outcome would make that group too likely.
Reset between full trials, not between dependent stages
For draws without replacement from R1, R2, R3, B1, B2, select two distinct labels. The first is unavailable for the second draw. Restore all five labels only after that complete two-draw trial.
Record trials and successes, divide to estimate the probability, and compare with theory. Report both sampling variation and any mismatch between simulation assumptions and the real experiment.
A common trap
Use independent fixed colour probabilities for draws without replacement.
Instead: Remove the chosen label within a trial; otherwise the model simulates replacement.
See the thinking
Read supplied digits
Use 0–4 = H and 5–9 = T. Read pairs 07, 26, 44, 91, 58, 30. Estimate P(exactly one head).
- The outcomes are HT, HT, HH, TH, TT, HH.
- Three of the six trials have exactly one head.
A without-replacement simulation
Describe one fair simulated trial for two draws without replacement from R1, R2, R3, B1, B2.
- Randomly order the five distinct labels with each order equally likely.
- Use the first two labels and record their colours.
- Restore the full list before the next trial.
Check your understanding
In the equal-three-outcome digit mapping, what happens to digit 9?
Show answer
Reject it and draw again.
Digits 0–8 form three equally sized groups; 9 would unbalance them.
Your turn
6 questions · 13 marks · Work at your own pace.
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 6/10 and T 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.
- 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
.
More trials help stability, but the model must also fit the real rules.
0.396; a finite simulation still varies randomly.
Curriculum reference & lesson scope
Australian Curriculum v9 / SA Curriculum Mathematics
Year 9 compound-event sample spaces, probabilities, experiments and simulations.
Official curriculum source ↗