The idea, step by step
Match the random mechanism
Two fair coins can be simulated by four equally likely labels HH,HT,TH,TT. Two fair dice require two independent random integers from 1 to 6. Choosing a sum uniformly from 2 to 12 is wrong: sum 7 has six physical pairs, while sum 2 has only one.
Specify one complete trial
State the event, stage rules and reset rule before running. For replacement, restore before the next draw; without replacement, restore only after the full trial. Record every result, including failures. Report successes divided by trial count and compare with the model. The supplied table is original practice data, not a claim of an actual survey.
| Outcome | Count |
|---|---|
| HH | 5 |
| HT | 7 |
| TH | 6 |
| TT | 6 |
A common trap
Make every possible dice sum equally likely.
Instead: Generate each die fairly; equal-probability sums do not match two dice.
See the thinking
Worked example 1
Use the supplied simulation table to estimate P(exactly one head).
| Outcome | Count |
|---|---|
| HH | 5 |
| HT | 7 |
| TH | 6 |
| TT | 6 |
- HT and TH are successful.
- 7+6=13 successes out of 24.
- Theory is 2/4=1/2.
Worked example 2
Use random integers 1–4 to simulate blue with probability 1/4 and red with probability 3/4.
- Map 1 to blue.
- Map 2,3,4 to red.
- For independent stages generate a fresh integer each time.
Check your understanding
A simulation has 45 successes in 180 trials. Find its estimate.
Show answer
1/4=0.25.
45÷180=0.25.
Your turn
6 questions · 16 marks · Work at your own pace.
Read the questions online
- Map equally likely digits 0–9 to P(blue)=0.4 and P(red)=0.6.2 marks
- Two tokens are drawn without replacement in a trial. When is the full bag restored for repeated trials?2 marks
- Why does a uniform random integer from 2 to 12 not simulate a fair two-dice sum?3 marks
- Use the table to estimate P(at least one head).2 marks
Original practice simulation: 24 trials of two fair coins Outcome Count HH 5 HT 7 TH 6 TT 6 - A student deletes all TT trials as uninteresting. Explain the problem.3 marks
- Describe a reproducible simulation of two dice and sum 7, including a check.4 marks
Read the worked solutions
Equivalent correct methods earn the corresponding marks. Include units where asked.
- Map equally likely digits 0–9 to P(blue)=0.4 and P(red)=0.6.2 marks
Four labels versus six gives the required probabilities.
For example, 0–3 blue and 4–9 red. - Two tokens are drawn without replacement in a trial. When is the full bag restored for repeated trials?2 marks
Restoring between draws changes the experiment.
After both draws, before the next complete trial. - Why does a uniform random integer from 2 to 12 not simulate a fair two-dice sum?3 marks
The required outcome multiplicities differ.
It treats sums 2 and 7 equally although they have 1 and 6 physical pairs. - Use the table to estimate P(at least one head).2 marks
Original practice simulation: 24 trials of two fair coins Outcome Count HH 5 HT 7 TH 6 TT 6 Count HH,HT,TH:5+7+6=18.
18/24=3/4. - A student deletes all TT trials as uninteresting. Explain the problem.3 marks
All complete trials must be recorded.
Selective removal changes the denominator and distribution. - Describe a reproducible simulation of two dice and sum 7, including a check.4 marks
Each ordered pair is modelled fairly and the trial count is reported.
Generate two independent integers 1–6 each trial, count sum 7, divide by trial count, and compare with 6/36=1/6.
Curriculum reference & lesson scope
Australian Curriculum v9 and SA Curriculum
Complementary events; two-stage sample spaces; tables, trees and Venn diagrams; probabilities of combined outcomes; experimental versus theoretical results; compound-event simulations and describing results. Formal conditional probability is Year 10.
Official curriculum source ↗