<p><strong>For Problems 16–18:</strong> Two fair dice are rolled. Let \(P(A_i) > 0\) denote the event that the sum of the faces of the dice is divisible by \(i\).</p><p>For which one of the following pairs \((i, j)\) are the events \(A_i\) and \(A_j\) independent?</p>
Step-by-Step Solution
Key Concept: Two events are independent if P(A_i ∩ A_j) = P(A_i) · P(A_j). For dice sums, we must count outcomes where the sum is divisible by both i and j (i.e., by lcm(i,j)), then verify the independence condition.
<p><strong>Step 1:</strong> Total outcomes when rolling two dice = 36.</p><p><strong>Step 2:</strong> For each pair (i,j), calculate P(A_i), P(A_j), and P(A_i ∩ A_j) by counting outcomes where the sum is divisible by i, j, and lcm(i,j) respectively.</p><p><strong>Step 3:</strong> Check independence condition: P(A_i ∩ A_j) = P(A_i) · P(A_j)?</p><p><strong>Example analysis for common pairs:</strong></p><p>• <strong>(3,4):</strong> Sums divisible by 3: {3,6,9,12} → 12 outcomes; P(A_3) = 12/36 = 1/3. Sums divisible by 4: {4,8,12} → 9 outcomes; P(A_4) = 9/36 = 1/4. Sums divisible by lcm(3,4)=12: only {12} → 1 outcome; P(A_3 ∩ A_4) = 1/36. Check: (1/3)·(1/4) = 1/12 ≠ 1/36. <strong>Not independent.</strong></p><p>• <strong>(2,3):</strong> P(A_2) = 18/36 = 1/2. P(A_3) = 12/36 = 1/3. Sums divisible by lcm(2,3)=6: {6,12} → 6 outcomes; P(A_2 ∩ A_3) = 6/36 = 1/6. Check: (1/2)·(1/3) = 1/6 ✓ <strong>Independent!</strong></p><p><strong>∴ Answer: C (for pair (2,3) or equivalent independent pair)</strong></p>
Correct Answer: C