<p>For any three events \(A\), \(B\) and \(C\) defined on the sample space \(R\) such that \(B \subset C\) and \(P(A) \neq 0\), \(P(B/A) \leq P(C/A)\).</p>
Step-by-Step Solution
Key Concept: Since B ⊂ C, we have B ∩ A ⊆ C ∩ A. By the monotonicity property of probability, P(B ∩ A) ≤ P(C ∩ A). Dividing both sides by P(A) > 0 preserves the inequality, yielding P(B/A) ≤ P(C/A).
<p><strong>Step 1:</strong> Recall that conditional probability is defined as P(B/A) = P(B ∩ A)/P(A) when P(A) ≠ 0.</p><p><strong>Step 2:</strong> Given that B ⊂ C, we have B ∩ A ⊆ C ∩ A (the intersection preserves the subset relation).</p><p><strong>Step 3:</strong> By the monotonicity property of probability: if B ∩ A ⊆ C ∩ A, then P(B ∩ A) ≤ P(C ∩ A).</p><p><strong>Step 4:</strong> Divide both sides by P(A) > 0 (given that P(A) ≠ 0). Since we're dividing by a positive number, the inequality is preserved:</p><p>P(B ∩ A)/P(A) ≤ P(C ∩ A)/P(A)</p><p><strong>Step 5:</strong> By definition of conditional probability: P(B/A) ≤ P(C/A)</p><p>∴ <strong>Answer: A (The statement is TRUE)</strong></p>
Correct Answer: A