Mathematics Advanced • Year 12 • Module 5 • Lesson 2
Probability Rules
Past-paper style: addition, multiplication and complement rules, including a 7-mark proof and modelling response.
1. Short-answer questions
1.1 Events A and B satisfy P(A) = 0.4, P(B) = 0.55 and P(A ∪ B) = 0.7. Find P(A ∩ B), and hence find P(A′ ∩ B′). 2 marks Band 3
1.2 A factory makes 1 000 light bulbs per day. The probability that a single bulb is defective is 0.015, independently of others. Find P(at least one defective bulb in a day's production). Give your answer to 4 decimal places. 3 marks Band 3-4
1.3 A fair coin is flipped until either two heads appear in total, or four flips have been made — whichever comes first.
(a) List all possible outcome sequences and the probability of each.
(b) Find P(the experiment ends with two heads). 4 marks Band 4
2. Extended response
2.1 Let A and B be two events in a probability space.
(a) Prove from the axioms / addition rule that P(A ∩ B) ≤ min(P(A), P(B)). State the geometric meaning in terms of a Venn diagram, and describe exactly when equality P(A ∩ B) = P(A) is achieved.
(b) Hence prove the Bonferroni-style bound: for any two events,
P(A ∩ B) ≥ P(A) + P(B) − 1.
(c) Apply both bounds to events with P(A) = 0.7 and P(B) = 0.8 to find the smallest and largest possible values of P(A ∩ B). 7 marks Band 5-6
Explicit marking criteria
Part (a) — 2 marks
• 1 mark — argues A ∩ B ⊆ A and A ∩ B ⊆ B, using monotonicity of probability to deduce P(A ∩ B) ≤ P(A) and P(A ∩ B) ≤ P(B).
• 1 mark — states equality P(A ∩ B) = P(A) iff A ⊆ B (Venn picture: circle A entirely inside circle B).
Part (b) — 3 marks
• 1 mark — starts from the addition rule: P(A ∪ B) = P(A) + P(B) − P(A ∩ B).
• 1 mark — uses P(A ∪ B) ≤ 1 to deduce P(A) + P(B) − P(A ∩ B) ≤ 1.
• 1 mark — rearranges to P(A ∩ B) ≥ P(A) + P(B) − 1 with a closing sentence.
Part (c) — 2 marks — applies both bounds: P(A ∩ B) ≥ 0.7 + 0.8 − 1 = 0.5 and P(A ∩ B) ≤ min(0.7, 0.8) = 0.7. So 0.5 ≤ P(A ∩ B) ≤ 0.7. 1 mark per bound.
Your response:
Stuck on (b)? Rearrange the addition rule for P(A ∩ B), then use the fact that any probability is ≤ 1.How did this worksheet feel?
What I'll revisit before next class:
1.1 — Solving for P(A ∩ B) and P(A′ ∩ B′) (2 marks)
Sample response. From the addition rule, P(A ∪ B) = P(A) + P(B) − P(A ∩ B), so 0.7 = 0.4 + 0.55 − P(A ∩ B), giving P(A ∩ B) = 0.95 − 0.7 = 0.25. Then by De Morgan, P(A′ ∩ B′) = P((A ∪ B)′) = 1 − P(A ∪ B) = 1 − 0.7 = 0.3.
Marking notes. 1 mark — correctly rearranges addition rule to find P(A ∩ B) = 0.25. 1 mark — applies De Morgan or complement reasoning to give P(A′ ∩ B′) = 0.3.
1.2 — At least one defective bulb (3 marks)
Sample response. Defectiveness of each bulb is independent. P(one bulb non-defective) = 0.985. P(all 1 000 non-defective) = (0.985)¹⁰⁰⁰. Taking logs: 1 000 × ln(0.985) ≈ 1 000 × (−0.015114) = −15.114, so (0.985)¹⁰⁰⁰ ≈ e⁻¹⁵·¹¹⁴ ≈ 2.72 × 10⁻⁷.
Therefore P(at least one defective) = 1 − 2.72 × 10⁻⁷ ≈ 0.9999997 (very close to 1, to 4 d.p. is 1.0000).
Marking notes. 1 mark — identifies complement strategy. 1 mark — computes (0.985)¹⁰⁰⁰ correctly using independence. 1 mark — final value with comment that "at least one" is virtually certain across 1 000 trials. Acceptable: stating 0.9999997 to 7 d.p. or 1.0000 to 4 d.p. with explanation.
1.3 — Coin flips, stop at 2H or 4 flips (4 marks)
Sample response.
(a) Outcome sequences and their probabilities:
- HH (stops on flip 2, 2 heads): P = (1/2)² = 1/4
- HTH (stops on flip 3, 2 heads): P = (1/2)³ = 1/8
- THH (stops on flip 3, 2 heads): P = (1/2)³ = 1/8
- HTTH (stops on flip 4, 2 heads): P = (1/2)⁴ = 1/16
- THTH (stops on flip 4, 2 heads): P = (1/2)⁴ = 1/16
- TTHH (stops on flip 4, 2 heads): P = (1/2)⁴ = 1/16
- HTTT, THTT, TTHT, TTTH, TTTT (4 flips, fewer than 2 heads): each P = 1/16
(Check that all sequence probabilities sum to 1: 1/4 + 2(1/8) + 3(1/16) + 5(1/16) = 4/16 + 4/16 + 3/16 + 5/16 = 16/16 ✓.)
(b) P(ends with 2 heads) = 1/4 + 2(1/8) + 3(1/16) = 4/16 + 4/16 + 3/16 = 11/16.
Marking notes. 1 mark — identifies the "stops on flip 2" case correctly (HH only). 1 mark — identifies the "stops on flip 3" cases (HTH, THH). 1 mark — identifies the "stops on flip 4 with 2 heads" cases (HTTH, THTH, TTHH). 1 mark — sums correctly to 11/16. Common error: including 4-flip sequences with fewer than 2 heads in the success count.
2.1 — Bounds on P(A ∩ B) (7 marks): sample Band-6 response with annotations
Sample Band-6 response.
Part (a). By definition, A ∩ B is the set of outcomes in both A and B, so every element of A ∩ B is also an element of A. That is, A ∩ B ⊆ A. By monotonicity of probability, P(A ∩ B) ≤ P(A). The same argument with A and B swapped gives P(A ∩ B) ≤ P(B). Therefore
P(A ∩ B) ≤ min(P(A), P(B)). [1 mark — subset/monotonicity argument.]
Equality P(A ∩ B) = P(A) holds iff A ⊆ B: every outcome in A is also in B, so the intersection equals A itself. On a Venn diagram, the circle A sits entirely inside circle B. [1 mark — equality condition with Venn interpretation.]
Part (b). From the addition rule,
P(A ∪ B) = P(A) + P(B) − P(A ∩ B). [1 mark — starts from addition rule.]
Because P(A ∪ B) is a probability, P(A ∪ B) ≤ 1. Substituting,
P(A) + P(B) − P(A ∩ B) ≤ 1. [1 mark — applies probability ≤ 1.]
Rearranging,
P(A ∩ B) ≥ P(A) + P(B) − 1.
This is the Bonferroni lower bound; equality holds when P(A ∪ B) = 1 (the events cover the whole sample space). [1 mark — rearrangement and closing comment.]
Part (c). Apply both bounds with P(A) = 0.7, P(B) = 0.8:
P(A ∩ B) ≥ 0.7 + 0.8 − 1 = 0.5 (Bonferroni lower bound). [1 mark]
P(A ∩ B) ≤ min(0.7, 0.8) = 0.7 (subset upper bound). [1 mark]
So 0.5 ≤ P(A ∩ B) ≤ 0.7. The lower bound is achieved when A ∪ B = S; the upper bound when A ⊆ B. ▮
Total: 7/7.
Band descriptors for marker.
Band 3: States the addition rule and substitutes; finds part (c) numerically but without justification from (a) or (b). ≈ 2-3 marks.
Band 4: Proves the upper bound in (a) using subset reasoning; attempts (b) but does not link P(A ∪ B) ≤ 1 cleanly; (c) correct. ≈ 4-5 marks.
Band 5: Both proofs complete; (c) gives both bounds but does not explain when each is achieved. ≈ 5-6 marks.
Band 6: Full proofs with named conditions for equality, Venn-diagram interpretation, and explicit statement of when each bound is tight. 7/7.