ZIMSEC A Level · Statistics · Form 5
Rules
- P(A′) = 1 − P(A).
- Addition: P(A ∪ B) = P(A) + P(B) − P(A ∩ B). Mutually exclusive events: P(A ∩ B) = 0.
- Independent events: P(A ∩ B) = P(A) × P(B).
- Conditional probability: P(A | B) = P(A ∩ B) ÷ P(B).
Tree diagrams and Venn diagrams
Multiply along branches; add the end results. Venn diagrams organise overlapping events.
Worked example: 60% of learners take Mathematics, 40% take Physics and 30% take both. Find P(Physics | Mathematics) and decide whether the events are independent.
P(P | M) = 0.30 ÷ 0.60 = 0.5
P(M) × P(P) = 0.6 × 0.4 = 0.24 ≠ 0.30 → not independent
Bayes' theorem (reverse conditional probability)
P(A | B) = P(B | A)P(A) ÷ P(B), with P(B) found from a tree diagram.
Worked example: 2% of a population has a disease. A test detects it 95% of the time but gives a false positive 5% of the time. Find P(disease | positive).
P(positive) = 0.02 × 0.95 + 0.98 × 0.05 = 0.019 + 0.049 = 0.068
P(disease | positive) = 0.019 ÷ 0.068 ≈ 0.279
Check yourself
- P(A) = 0.5, P(B) = 0.3, P(A ∪ B) = 0.65. Find P(A ∩ B).
- Are A and B in the previous question independent?
- Two fair dice are thrown. Find P(total = 7).
Show answers
- 0.5 + 0.3 − 0.65 = 0.15
- Yes: 0.5 × 0.3 = 0.15 = P(A ∩ B).
- 6/36 = 1/6
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