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.
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