Question:
<p>\(P(A)=0.5\), \(P(B)=0.3\), \(A\) and \(B\) mutually exclusive. \(P(\text{exactly one of }A,B)=\) <em>[JEE Advanced 2015]</em></p>
Step-by-Step Solution
Key Concept: Mutually exclusive: P(A\capB) = 0. P(exactly one) = P(A) + P(B) - 2P(A\capB) = P(A) + P(B).
<p>Since $A$ and $B$ are mutually exclusive, $P(A\cap B)=0$.</p><p>$P(\text{exactly one}) = P(A\cup B) - P(A\cap B) = P(A)+P(B) = 0.5+0.3 = 0.8$</p><p>Equivalently: $P(\text{exactly one}) = P(A)+P(B)-2P(A\cap B) = 0.8-0 = 0.8$. Answer: B ✓</p>
Correct Answer: B