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.