Mathematics Advanced • Year 12 • Module 5 • Lesson 2
Probability Rules
Build fluency with the addition rule, multiplication rule and complement, including "at least one" reasoning.
1. Quick recall
Answer each question in the space provided. 1 mark each
Q1.1 Complete each rule in the form used on the formula sheet:
Addition rule: P(A ∪ B) = _______________________________________
Multiplication rule (general): P(A ∩ B) = _________________________
Multiplication rule (independent): P(A ∩ B) = _______________________
Q1.2 The complement rule and "at least one" rule:
P(A′) = ______________ P(at least one) = 1 − P(_____________)
When is "at least one" the right method to reach for? ____________________________________________
Q1.3 Two events have P(A) = 0.3 and P(B) = 0.6. State the value of P(A ∪ B) if A and B are mutually exclusive: ______. Could P(A ∪ B) ever be smaller than P(B)? Yes / No (circle) Why? ________________
2. Worked example — defective items from two machines
Problem. A factory has two machines. Machine A produces 60% of items with a 2% defect rate. Machine B produces 40% of items with a 5% defect rate. An item is selected at random. Find P(defective).
Step 1 — Identify partitions and conditional probabilities.
P(A) = 0.6, P(B) = 0.4 (the two machines partition the items)
P(def | A) = 0.02, P(def | B) = 0.05
Reason: each item comes from exactly one machine, so we condition on the source.
Step 2 — Multiply along each route to find joint probabilities.
P(A ∩ def) = P(A) × P(def | A) = 0.6 × 0.02 = 0.012
P(B ∩ def) = P(B) × P(def | B) = 0.4 × 0.05 = 0.020
Reason: multiplication rule P(A ∩ B) = P(A) · P(B | A).
Step 3 — Add the two mutually exclusive routes (total probability).
P(def) = 0.012 + 0.020 = 0.032
Reason: an item can only come from A or B (not both), so routes don't overlap — addition rule for mutually exclusive events.
Conclusion. P(defective) = 0.032 (3.2%).
3. Faded example — at least one six in three rolls
A fair die is rolled three times. Find P(at least one six). Fill in each blank. 4 marks
Step 1 — Define the complement. "At least one six" is the complement of "______________________".
Step 2 — Probability of no six on one roll. P(not 6) = ______ / 6 = ______
Step 3 — Use independence across three rolls.
P(no six on all three) = P(not 6)³ = ( ______ )³ = ______ / ______
Step 4 — Subtract from 1.
P(at least one six) = 1 − ______ / ______ = ______ / ______
Conclusion. P(at least one six) = ____________ ≈ ______ (3 d.p.)
4. Graduated practice
Show one line of working and a final answer as a simplified fraction or 3 d.p. decimal.
Foundation — direct rule applications (4 questions)
| Q | Question | Working | Answer |
|---|---|---|---|
| 4.1 1 | P(A) = 0.5, P(B) = 0.4, P(A ∩ B) = 0.2. Find P(A ∪ B). | ||
| 4.2 1 | P(A) = 0.3, P(B) = 0.6, A and B mutually exclusive. Find P(A ∪ B). | ||
| 4.3 1 | P(A) = 0.4, P(B | A) = 0.25. Find P(A ∩ B). | ||
| 4.4 1 | P(A) = 0.7. Find P(A′). |
Standard — typical HSC difficulty (6 questions)
4.5 A box has 6 red and 4 green balls. Two balls are drawn without replacement. Find P(red then green). 2 marks
4.6 A bag has 8 black and 12 white counters. Two are drawn with replacement. Find P(both black). 2 marks
4.7 P(A) = 0.6, P(B) = 0.5, P(A ∩ B) = 0.3. Find: (a) P(A ∪ B); (b) P(A′ ∩ B) (i.e. B occurs but not A). 2 marks
4.8 A fair coin is flipped 5 times. Find P(at least one head) by complement. 2 marks
4.9 Four people each randomly guess a number from 1 to 10 (independently). Find P(at least one correct guess). 2 marks
4.10 A bag has 4 red, 5 blue and 6 green marbles. Two marbles are drawn without replacement. Find P(both red). 2 marks
Extension — combine rules (2 questions)
4.11 Show that for any two events A and B, P(A ∩ B) ≤ min(P(A), P(B)). State exactly when equality P(A ∩ B) = P(A) holds. 3 marks
4.12 A multiple-choice test has 10 questions, each with 4 options. A student guesses every answer. Find P(at least one correct). Then state the smallest n (number of questions) for which P(at least one correct) exceeds 0.99. 3 marks
5. Self-check the easy 3
Tick the first three once you've verified your method works.
How did this worksheet feel?
What I'll revisit before next class:
Q1.1 — Rule recall
Addition: P(A ∪ B) = P(A) + P(B) − P(A ∩ B). Multiplication (general): P(A ∩ B) = P(A) × P(B | A). Multiplication (independent): P(A ∩ B) = P(A) × P(B).
Q1.2 — Complement and at least one
P(A′) = 1 − P(A); P(at least one) = 1 − P(none). Use it whenever the question contains the phrase "at least one" or a long list of "exactly k" cases.
Q1.3 — Mutually exclusive bound
P(A ∪ B) = 0.3 + 0.6 = 0.9. No — P(A ∪ B) ≥ max(P(A), P(B)) = 0.6 because B ⊆ A ∪ B, so the union always contains at least as much as either event.
Q3 — Faded example (at least one six in 3 rolls)
Step 1: complement of "no sixes in any of the three rolls".
Step 2: P(not 6) = 5/6.
Step 3: P(no six in 3 rolls) = (5/6)³ = 125/216.
Step 4: P(at least one six) = 1 − 125/216 = 91/216.
Conclusion: 91/216 ≈ 0.421.
Q4.1 — Addition rule
P(A ∪ B) = 0.5 + 0.4 − 0.2 = 0.7.
Q4.2 — Mutually exclusive addition
P(A ∪ B) = 0.3 + 0.6 = 0.9.
Q4.3 — Multiplication rule
P(A ∩ B) = 0.4 × 0.25 = 0.10.
Q4.4 — Complement
P(A′) = 1 − 0.7 = 0.3.
Q4.5 — Red then green (without replacement)
P(red first) = 6/10. After removing one red, P(green | red first) = 4/9. So P = (6/10)(4/9) = 24/90 = 4/15.
Q4.6 — Both black (with replacement)
With replacement, draws are independent: P = (8/20)(8/20) = 64/400 = 4/25.
Q4.7 — Mixed
(a) P(A ∪ B) = 0.6 + 0.5 − 0.3 = 0.8. (b) P(A′ ∩ B) = P(B) − P(A ∩ B) = 0.5 − 0.3 = 0.2.
Q4.8 — At least one head in 5 flips
P(no heads) = (1/2)⁵ = 1/32. P(at least one head) = 1 − 1/32 = 31/32 ≈ 0.969.
Q4.9 — At least one correct guess
P(one person wrong) = 9/10. P(all four wrong) = (9/10)⁴ = 0.6561. So P(at least one correct) = 1 − 0.6561 = 0.3439.
Q4.10 — Both red (without replacement)
Total marbles = 15. P = (4/15)(3/14) = 12/210 = 2/35.
Q4.11 — Proof P(A ∩ B) ≤ min(P(A), P(B))
A ∩ B ⊆ A, so P(A ∩ B) ≤ P(A). Similarly A ∩ B ⊆ B, so P(A ∩ B) ≤ P(B). Therefore P(A ∩ B) ≤ min(P(A), P(B)). Equality P(A ∩ B) = P(A) holds when A ⊆ B (every outcome in A is also in B), so the intersection equals A entirely.
Q4.12 — MCQ guessing
P(wrong on one question) = 3/4. P(all 10 wrong) = (3/4)¹⁰ ≈ 0.0563. So P(at least one correct) ≈ 1 − 0.0563 = 0.944.
For P(at least one correct) > 0.99 we need (3/4)ⁿ < 0.01 ⇒ n log(3/4) < log(0.01) ⇒ n > log(0.01)/log(0.75) ≈ 16.0. Smallest integer n = 17.