Probability Theory and Random Variables
Domain 3: Probability Theory and Random Variables
This domain tests the axiomatic building of probability, conditional probability and Bayes' theorem, and the expectation and variance of a random variable defined on a sample space. Questions are short but hinge on correctly identifying the conditioning event.
Core rules
- 0 ≤ P(A) ≤ 1, P(S) = 1, P(A′) = 1 − P(A).
- Addition: P(A ∪ B) = P(A) + P(B) − P(A ∩ B); the last term vanishes only for mutually exclusive events.
- Conditional: P(A|B) = P(A ∩ B)/P(B) for P(B) > 0; multiplication P(A ∩ B) = P(A)P(B|A).
- Independence: P(A ∩ B) = P(A)P(B), equivalently P(A|B) = P(A).
- Total probability: P(B) = Σ P(A_i)P(B|A_i) over an exhaustive, mutually exclusive partition.
- Bayes: P(A_i|B) = P(A_i)P(B|A_i) / Σ P(A_j)P(B|A_j).
- Discrete variable: pmf p(x) = P(X = x), Σp(x) = 1; E(X) = Σx p(x); Var(X) = E(X²) − [E(X)]².
- Linear rules: E(aX + b) = aE(X) + b and Var(aX + b) = a²Var(X), so the SD scales by |a| only.
Worked example
Urn I holds 2 white and 3 red balls; Urn II holds 4 white and 1 red. An urn is chosen at random and one ball drawn; it is white. Then P(I)P(W|I) = 0.5 × 0.4 = 0.20 and P(II)P(W|II) = 0.5 × 0.8 = 0.40, so P(W) = 0.60. P(I|W) = 0.20/0.60 = 1/3. Note the reversal: the forward probability 0.4 becomes 1/3 once the evidence is used, which is exactly the correction Bayes' theorem performs.
Traps
- Adding probabilities of events that can occur together.
- Confusing mutually exclusive with independent; for events of positive probability the two cannot both hold.
- Writing P(A|B) where P(B|A) is meant - the classic error in urn and disease-testing questions.
- Forgetting to change the denominator when drawing without replacement.
- Using Var(aX) = aVar(X) instead of a²Var(X).