Probability: Conditional Probability, Independence, Bayes and Distributions
This domain tests the algebra of events, conditional probability, independence, the total probability theorem and Bayes' theorem (usually through a tree diagram or a contingency table), and the binomial and normal distributions with their parameters and tables.
Basic rules
- P(A ∪ B) = P(A) + P(B) − P(A ∩ B); the last term vanishes only for mutually exclusive events.
- P(Ā) = 1 − P(A).
- P(A|B) = P(A ∩ B)/P(B) for P(B) > 0.
- A and B are independent iff P(A ∩ B) = P(A)·P(B), equivalently P(A|B) = P(A). Independence must be justified, never assumed from the wording alone.
Total probability and Bayes
If A₁, A₂ form a partition, P(B) = P(A₁)P(B|A₁) + P(A₂)P(B|A₂), and P(Aᵢ|B) = P(Aᵢ)P(B|Aᵢ)/P(B). The denominator is the total probability already computed; a tree with branches for the causes and leaves for the result makes the arithmetic transparent.
Binomial and normal
- X ~ B(n,p): P(X = k) = C(n,k)·pᵏ·(1−p)ⁿ⁻ᵏ, with μ = np and σ = √(np(1−p)). It models n independent repetitions with constant p.
- X ~ N(μ,σ): standardise z = (x − μ)/σ (divide by σ, never σ²) and read Φ from the table; P(Z > a) = 1 − Φ(a), P(Z < −a) = 1 − Φ(a), P(a < Z < b) = Φ(b) − Φ(a).
Worked example
Machine A produces 60% of the bolts with 2% defective; machine B produces 40% with 5% defective. P(defective) = 0.6·0.02 + 0.4·0.05 = 0.012 + 0.020 = 0.032. Given that a bolt is defective, P(A | defective) = 0.012/0.032 = 0.375.
Traps that cost marks
- Confusing P(A|B) with P(B|A) - the classic error with diagnostic tests.
- Using P(A ∩ B) = P(A)P(B) when independence has not been established.
- Inserting the variance instead of the standard deviation in z = (x − μ)/σ.
- Omitting the combinatorial factor C(n,k) in the binomial formula, or using Φ(−a) = Φ(a) instead of 1 − Φ(a).