Probability
Conditional Probability
Grade 12
Question:
<p>Let \(A\) and \(E\) be any two events with positive probabilities.<br><strong>Statement-1:</strong> \(P(E/A) \geq P(A/E)P(E)\)<br><strong>Statement-2:</strong> \(P(A/E) \geq P(A \cap E)\).</p>
<p>Both the statements are false.</p>
<p>Both the statements are true.</p>
<p>Statement-1 is true, Statement-2 is false.</p>
<p>Statement-1 is false, Statement-2 is true.</p>
Step-by-Step Solution
Key Concept: Use the definition of conditional probability P(A|B) = P(A∩B)/P(B) to rewrite statements algebraically, then compare them using the constraint that probabilities are at most 1.
<p><strong>Step 1: Analyze Statement-1:</strong> P(E|A) ≥ P(A|E)·P(E)</p><p>LHS = P(E|A) = P(A∩E)/P(A)</p><p>RHS = P(A|E)·P(E) = [P(A∩E)/P(E)]·P(E) = P(A∩E)</p><p>So Statement-1 claims: P(A∩E)/P(A) ≥ P(A∩E)</p><p>This simplifies to: P(A∩E) ≥ P(A∩E)·P(A), or equivalently 1 ≥ P(A)</p><p>✓ Always TRUE since P(A) ≤ 1</p><p><strong>Step 2: Analyze Statement-2:</strong> P(A|E) ≥ P(A∩E)</p><p>LHS = P(A|E) = P(A∩E)/P(E)</p><p>Statement claims: P(A∩E)/P(E) ≥ P(A∩E)</p><p>This requires: 1/P(E) ≥ 1, or P(E) ≤ 1</p><p>This is NOT always true. Counterexample: If P(E) = 0.5 and P(A∩E) = 0.3, then P(A|E) = 0.6 > 0.3 ✓, but if P(A∩E) = 0.6, then P(A|E) = 1.2 > 0.6 ✗ (probability cannot exceed 1)</p><p>✗ Statement-2 is FALSE</p><p><strong>Step 3: Conclusion</strong></p><p>Statement-1 is TRUE, Statement-2 is FALSE</p><p>∴ Answer: <strong>B</strong> (Statement-1 is true; Statement-2 is false)</p>
Correct Answer: B