Start learning →

Events $A$ and $B$ are mutually exclusive with $P(A) = 0.4$ and $P(B) = 0.3$. Find $P(A \cup B)$.

A$0.12$
B$0.70$
C$1.00$
D$0.58$
Answer & Solution
Correct answer: B. $0.70$
Mutually exclusive events cannot occur together, so $P(A \cap B) = 0$. The general addition rule simplifies: $P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.4 + 0.3 - 0 = 0.7$. Option A ($0.12$) is the product $P(A) \cdot P(B)$, which is the joint probability for *independent* events, not mutually exclusive ones. **The distinction matters:** mutually exclusive means $P(A \cap B) = 0$ independent means $P(A \cap B) = P(A) \cdot P(B)$. These two assumptions cannot both hold unless one event has zero probability.
Practise this question with Quill — GMAT →

Keep practising Geometry: Circles

Browse all Geometry: Circles questions for GMAT, or explore more Mathematics topics.

Start a Geometry: Circles lesson →