If $P(A) = 0.4$ and $P(B) = 0.5$ and $P(A \cap B) = 0.2$, then $P(A \cup B)$ is

1. Apply the addition rule: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$. 2. Substitute: $P(A \cup B) = 0.4 + 0.5 - 0.2 = 0.7$. 3. Sanity check: $P(A \cup B)$ must be $\leq 1$ (it's a probability) AND $\geq$ each individual $P(A), P(B)$. Both satisfied: $0.7 \geq 0.5 \geq 0.4$. ✓ 4. Option A forgets to subtract $P(A \cap B)$ (gives $0.9$, also valid). Option C is just $P(A \cap B)$ confusing union with intersection. Option D ($1.1$) is impossible — probabilities never exceed 1. _Source: syllabus Mathematics, Ch 14, §14.2.3 (Addition rule, numerical examples), p. 11._