Answer: Var(X) + Var(Y) + 2Cov(X,Y).
- A Var(X) + Var(Y), the independent-case formula
- B Var(X) + Var(Y) + 2Cov(X,Y)
- C Var(X) × Var(Y), an incorrect multiplicative form
- D Cov(X,Y) alone, without the individual variance terms
Correct answer: B. Var(X) + Var(Y) + 2Cov(X,Y)
Explanation: Expand: Var(X+Y) = E[(X+Y−μₓ−μᵧ)²] = E[(X−μₓ)²] + 2E[(X−μₓ)(Y−μᵧ)] + E[(Y−μᵧ)²] = Var(X) + 2Cov(X,Y) + Var(Y). When independent, Cov(X,Y)=0, reducing to Var(X)+Var(Y).
Venn diagram of the universal set U with events A and B, showing the intersection (A and B), the parts unique to each event, and the complement region outside both.
Concept context
Chance, events, conditional probability, and Bayes theorem