Australian Curriculum v9 / ACiQ Year 8 Mathematics - Unit 4 - Complementary and compound probability with simulation

Complementary and compound probability with simulation

Use complements, two-step sample spaces and simulations to calculate and estimate probabilities.

Updated 2026-07-26 - 12 min read

Complementary and compound probability with simulation is taught here as a connected set of decisions, not a list of facts. Work through the prerequisite recall, explicit models, carefully faded examples, misconception repairs and transfer task before using the target in Check, Practice, Review or Rapid Revision.

This note is designed to work with the guided lessons, curated practice, flashcards, Tutor context, Review and Rapid Revision for the same canonical target. The same three evidence checks are used throughout, so feedback can route a learner back to the precise idea that needs repair.

Use complementary probability

An event and its complement exhaust all outcomes, so P(not A) = 1 − P(A). This relationship must be selected from the quantities and conditions in the problem, then checked against the context.

A dependable reasoning routine

  1. Name the unknowns, units and constraints before calculating.
  2. Choose the relationship represented by use complementary probability and state why it applies.
  3. Keep exact values for as long as possible, show substitutions and preserve units through each step.
  4. Check the result by substitution, estimation, an alternative representation or the original context.

Repair: The complement includes every outcome outside the event, so not-6 includes 1, 2, 3, 4 and 5.

The repair matters because the shortcut may appear to work in one familiar example while failing when the method, representation, scale, constraint or accuracy requirement changes. Use the routine above to make the reasoning visible enough for another learner to verify.

Example 1.1

If P(rain) = 0.35, find P(no rain).

Step 1 - identify the governing idea: An event and its complement exhaust all outcomes, so P(not A) = 1 − P(A).

Step 2 - apply it to this evidence: Complementary probabilities sum to 1.

Result: 0.65

The relationship is visible in the working: Complementary probabilities sum to 1. Check the units and substitute or estimate where possible. A plausible-looking number is not enough unless it satisfies the original conditions.

Why the alternatives fail:

  • 0.35 — It does not agree with the required relationship: Complementary probabilities sum to 1.
  • 1.35 — It changes or misses a condition in the question. Reapply the relationship and retain the stated units or accuracy.
  • 0.175 — A substitution, estimate, ordering or unit check rejects this result; it does not reproduce the conditions in the prompt.

Example 1.2

A bag has a 3/8 chance of red. Find the chance of not red.

Step 1 - identify the governing idea: An event and its complement exhaust all outcomes, so P(not A) = 1 − P(A).

Step 2 - apply it to this evidence: 1 − 3/8 = 5/8.

Result: 5/8

The relationship is visible in the working: 1 − 3/8 = 5/8. Check the units and substitute or estimate where possible. A plausible-looking number is not enough unless it satisfies the original conditions.

Why the alternatives fail:

  • 3/8 — It does not agree with the required relationship: 1 − 3/8 = 5/8.
  • 1/8 — It changes or misses a condition in the question. Reapply the relationship and retain the stated units or accuracy.
  • 11/8 — A substitution, estimate, ordering or unit check rejects this result; it does not reproduce the conditions in the prompt.

Example 1.3

If P(A) = 72%, what is P(A does not occur)?

Step 1 - identify the governing idea: An event and its complement exhaust all outcomes, so P(not A) = 1 − P(A).

Step 2 - apply it to this evidence: 100% − 72% = 28%.

Result: 28%

The relationship is visible in the working: 100% − 72% = 28%. Check the units and substitute or estimate where possible. A plausible-looking number is not enough unless it satisfies the original conditions.

Why the alternatives fail:

  • 72% — It does not agree with the required relationship: 100% − 72% = 28%.
  • 172% — It changes or misses a condition in the question. Reapply the relationship and retain the stated units or accuracy.
  • 36% — A substitution, estimate, ordering or unit check rejects this result; it does not reproduce the conditions in the prompt.

Build two-step sample spaces

List ordered outcomes systematically using a table or tree, then count or add probabilities for the desired event. This relationship must be selected from the quantities and conditions in the problem, then checked against the context.

A dependable reasoning routine

  1. Name the unknowns, units and constraints before calculating.
  2. Choose the relationship represented by build two-step sample spaces and state why it applies.
  3. Keep exact values for as long as possible, show substitutions and preserve units through each step.
  4. Check the result by substitution, estimation, an alternative representation or the original context.

Repair: When every first-step outcome can pair with every second-step outcome, the number of ordered outcomes multiplies.

The repair matters because the shortcut may appear to work in one familiar example while failing when the method, representation, scale, constraint or accuracy requirement changes. Use the routine above to make the reasoning visible enough for another learner to verify.

Example 2.1

A coin is tossed and a six-sided die rolled. How many equally likely ordered outcomes exist?

Step 1 - identify the governing idea: List ordered outcomes systematically using a table or tree, then count or add probabilities for the desired event.

Step 2 - apply it to this evidence: There are 2 coin outcomes × 6 die outcomes.

Result: 12

The relationship is visible in the working: There are 2 coin outcomes × 6 die outcomes. Check the units and substitute or estimate where possible. A plausible-looking number is not enough unless it satisfies the original conditions.

Why the alternatives fail:

  • 8 — It does not agree with the required relationship: There are 2 coin outcomes × 6 die outcomes.
  • 6 — It changes or misses a condition in the question. Reapply the relationship and retain the stated units or accuracy.
  • 36 — A substitution, estimate, ordering or unit check rejects this result; it does not reproduce the conditions in the prompt.

Example 2.2

Two fair coins are tossed. What is P(exactly one head)?

Step 1 - identify the governing idea: List ordered outcomes systematically using a table or tree, then count or add probabilities for the desired event.

Step 2 - apply it to this evidence: HT and TH are 2 of the 4 equally likely ordered outcomes.

Result: 1/2

The relationship is visible in the working: HT and TH are 2 of the 4 equally likely ordered outcomes. Check the units and substitute or estimate where possible. A plausible-looking number is not enough unless it satisfies the original conditions.

Why the alternatives fail:

  • 1/4 — It does not agree with the required relationship: HT and TH are 2 of the 4 equally likely ordered outcomes.
  • 3/4 — It changes or misses a condition in the question. Reapply the relationship and retain the stated units or accuracy.
  • 1 — A substitution, estimate, ordering or unit check rejects this result; it does not reproduce the conditions in the prompt.

Example 2.3

A fair die is rolled twice. What is P(two sixes)?

Step 1 - identify the governing idea: List ordered outcomes systematically using a table or tree, then count or add probabilities for the desired event.

Step 2 - apply it to this evidence: Each roll has probability 1/6, so the ordered outcome (6,6) has probability 1/6 × 1/6.

Result: 1/36

The relationship is visible in the working: Each roll has probability 1/6, so the ordered outcome (6,6) has probability 1/6 × 1/6. Check the units and substitute or estimate where possible. A plausible-looking number is not enough unless it satisfies the original conditions.

Why the alternatives fail:

  • 1/12 — It does not agree with the required relationship: Each roll has probability 1/6, so the ordered outcome (6,6) has probability 1/6 × 1/6.
  • 1/6 — It changes or misses a condition in the question. Reapply the relationship and retain the stated units or accuracy.
  • 2/6 — A substitution, estimate, ordering or unit check rejects this result; it does not reproduce the conditions in the prompt.

Design and interpret probability simulations

A valid simulation matches the chance mechanism, uses many trials and estimates probability with relative frequency while retaining random variation. This relationship must be selected from the quantities and conditions in the problem, then checked against the context.

A dependable reasoning routine

  1. Name the unknowns, units and constraints before calculating.
  2. Choose the relationship represented by design and interpret probability simulations and state why it applies.
  3. Keep exact values for as long as possible, show substitutions and preserve units through each step.
  4. Check the result by substitution, estimation, an alternative representation or the original context.

Repair: Relative frequency is an estimate that tends to stabilise with more trials but remains subject to random variation.

The repair matters because the shortcut may appear to work in one familiar example while failing when the method, representation, scale, constraint or accuracy requirement changes. Use the routine above to make the reasoning visible enough for another learner to verify.

Example 3.1

How can a random digit generator simulate a 30% event?

Step 1 - identify the governing idea: A valid simulation matches the chance mechanism, uses many trials and estimates probability with relative frequency while retaining random variation.

Step 2 - apply it to this evidence: Three of ten equally likely digits give probability 3/10.

Result: Let digits 0, 1 and 2 represent success

The relationship is visible in the working: Three of ten equally likely digits give probability 3/10. Check the units and substitute or estimate where possible. A plausible-looking number is not enough unless it satisfies the original conditions.

Why the alternatives fail:

  • Let only digit 0 represent success — It does not agree with the required relationship: Three of ten equally likely digits give probability 3/10.
  • Use digits 0 to 4 as success — It changes or misses a condition in the question. Reapply the relationship and retain the stated units or accuracy.
  • Always output success three times first — A substitution, estimate, ordering or unit check rejects this result; it does not reproduce the conditions in the prompt.

Example 3.2

A simulation records 238 successes in 400 trials. Estimate probability.

Step 1 - identify the governing idea: A valid simulation matches the chance mechanism, uses many trials and estimates probability with relative frequency while retaining random variation.

Step 2 - apply it to this evidence: Relative frequency is 238 ÷ 400 = 0.595.

Result: 0.595

The relationship is visible in the working: Relative frequency is 238 ÷ 400 = 0.595. Check the units and substitute or estimate where possible. A plausible-looking number is not enough unless it satisfies the original conditions.

Why the alternatives fail:

  • 1.681 — It does not agree with the required relationship: Relative frequency is 238 ÷ 400 = 0.595.
  • 0.238 — It changes or misses a condition in the question. Reapply the relationship and retain the stated units or accuracy.
  • 59.5 — A substitution, estimate, ordering or unit check rejects this result; it does not reproduce the conditions in the prompt.

Example 3.3

Why repeat a simulation for more trials?

Step 1 - identify the governing idea: A valid simulation matches the chance mechanism, uses many trials and estimates probability with relative frequency while retaining random variation.

Step 2 - apply it to this evidence: Larger trial counts usually make relative frequency more stable.

Result: To reduce random fluctuation in the estimate

The relationship is visible in the working: Larger trial counts usually make relative frequency more stable. Check the units and substitute or estimate where possible. A plausible-looking number is not enough unless it satisfies the original conditions.

Why the alternatives fail:

  • To guarantee the exact probability — It does not agree with the required relationship: Larger trial counts usually make relative frequency more stable.
  • To change the theoretical model — It changes or misses a condition in the question. Reapply the relationship and retain the stated units or accuracy.
  • To remove the need for random outcomes — A substitution, estimate, ordering or unit check rejects this result; it does not reproduce the conditions in the prompt.

Retrieval check

Try these without looking back at the examples.

  1. A bag has a 3/8 chance of red. Find the chance of not red.
  2. Two fair coins are tossed. What is P(exactly one head)?
  3. A simulation records 238 successes in 400 trials. Estimate probability.

Answers

  1. 5/8 — 1 − 3/8 = 5/8.
  2. 1/2 — HT and TH are 2 of the 4 equally likely ordered outcomes.
  3. 0.595 — Relative frequency is 238 ÷ 400 = 0.595.

Transfer task

Find an unfamiliar example from school, daily life, a credible news source or another subject. Explain which of the three evidence checks applies. Complete the task, then audit your own response: identify the evidence used, the relationship applied, one plausible misconception and the final reasonableness check. If a peer could not reproduce your reasoning, add the missing step.

Sources