<p><em>A</em> and <em>B</em> are two independent events. <em>C</em> is an event in which exactly one of <em>A</em> or <em>B</em> occurs. Prove that
\[P(C) \geq P(A \cup B) \cdot P(\bar{A} \cap \bar{B}).\]</p>
Step-by-Step Solution
Key Concept: Express P(C) as P(A∩B̄) + P(Ā∩B) using independence, then recognize that P(A∪B) = 1 - P(Ā∩B̄) to convert the inequality into a form amenable to AM-GM or algebraic expansion.
<p><strong>Step 1:</strong> Express P(C) using independence.</p><p>Since C is the event where exactly one of A or B occurs:</p><p>P(C) = P(A∩B̄) + P(Ā∩B) = P(A)P(B̄) + P(Ā)P(B)</p><p>Let P(A) = p and P(B) = q. Then:</p><p>P(C) = p(1-q) + (1-p)q = p + q - 2pq</p><p><strong>Step 2:</strong> Express the RHS in terms of p and q.</p><p>P(A∪B) = P(A) + P(B) - P(A∩B) = p + q - pq (by independence)</p><p>P(Ā∩B̄) = P(Ā)P(B̄) = (1-p)(1-q) = 1 - p - q + pq</p><p>Therefore:</p><p>P(A∪B)·P(Ā∩B̄) = (p + q - pq)(1 - p - q + pq)</p><p><strong>Step 3:</strong> Prove the inequality algebraically.</p><p>Let x = p + q - pq and y = 1 - p - q + pq. Then x + y = 1.</p><p>We need: p + q - 2pq ≥ xy = x(1-x)</p><p>Since P(C) = p + q - 2pq and xy = x(1-x) where x = p + q - pq:</p><p>P(C) - xy = (p + q - 2pq) - (p + q - pq)(1 - p - q + pq)</p><p>= (p + q - 2pq) - (p + q - pq) + (p + q - pq)²</p><p>= -(pq) + (p + q - pq)² = (p - q)² + pq(1 - pq) ≥ 0</p><p>∴ P(C) ≥ P(A∪B)·P(Ā∩B̄)</p>
Correct Answer: Proof