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Year 10 · Probability · 18 min

Designing and interpreting simulations

Design fair probability simulations and interpret their variation without treating estimates as exact laws.

01 / LEARN

The idea, step by step

Make the simulation match the experiment

A simulation replaces an experiment with a random process that has the same relevant probabilities and rules. To simulate a 60% success chance using uniform random digits 0–9, let 0–5 mean success and 6–9 mean failure. Six of ten equally likely digits then represent success.

For two fair dice, generate two independent integers 1–6 and add them. Selecting one integer uniformly from 2–12 is not equivalent because real dice sums are not equally likely. For draws without replacement, update the available items instead of restarting the original bag.

Interpret estimates and variability

Relative frequency equals the number of observed event occurrences divided by the number of trials. It estimates the event probability; it does not redefine it. Finite runs vary, and a larger run is not guaranteed to be closer on every occasion.

Use enough trials, repeat runs and state the model assumptions. A simulated probability supports a prediction under those assumptions, not a claim that the real process has been measured or that an outcome is certain.

relative frequency=event counttrial count

A common trap

Expecting a larger simulation to produce the exact theoretical probability or always be closer.

Instead: Random variation remains. More trials generally stabilise estimates but do not guarantee a particular finite result.

02 / WORKED EXAMPLES

See the thinking

Example 1

A digit simulation

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

  1. Success digits are 0,4,3,1,5,2.
  2. There are 6 successes in 10 trials.
6/10=0.6.
Example 2

A flawed dice simulation

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

  1. The model assigns 1/11 to each sum.
  2. Actual sum 2 has probability 1/36 but sum 7 has 6/36.
Generate two independent die results instead.

Check your understanding

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

Show answer

0.63.

126/200=0.63; report it as an estimate.

03 / WORKSHEET

Your turn

6 questions · 15 marks · Work at your own pace.

Read the questions online
  1. Describe a random-digit simulation of a 30% chance.
    2 marks
  2. A simulation gives 47 events in 250 trials. Find relative frequency.
    2 marks
  3. Explain why generating one uniform integer 2–12 does not simulate two dice sums.
    2 marks
  4. Design a simulation of two counters drawn without replacement from 3 red and 2 blue.
    3 marks
  5. 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
  6. 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.

  1. 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.
  2. A simulation gives 47 events in 250 trials. Find relative frequency.
    2 marks

    47/250=0.188.

    0.188.
  3. 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.
  4. 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.
  5. 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.
  6. 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 ↗
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