Question:

<p><strong>Step 1: Find all possible sums and their probabilities.</strong> When two fair dice are rolled, sums range from 2 to 12. Total outcomes = 36.</p><p>P(A_i) = (number of ways sum is divisible by i)/36</p><p><strong>Step 2: Calculate P(A_i) for each relevant divisor i.</strong></p><p>• P(A_2) = 18/36 = 1/2 (sums: 2,4,6,8,10,12)</p><p>• P(A_3) = 12/36 = 1/3 (sums: 3,6,9,12)</p><p>• P(A_4) = 9/36 = 1/4 (sums: 4,8,12)</p><p>• P(A_5) = 7/36 (sums: 5,10)</p><p>• P(A_6) = 6/36 = 1/6 (sums: 6,12)</p><p>• P(A_7) = 6/36 = 1/6 (sums: 7)</p><p>• Other divisors yield different probabilities</p><p><strong>Step 3: Check independence for pairs (i,j).</strong> For independence: P(A_i ∩ A_j) = P(A_i)·P(A_j).</p><p><strong>Case 1: Pairs involving A_1</strong> A_1 is the entire sample space, P(A_1) = 1. Events A_1 and A_j are always independent for any j. This gives pairs: (1,2), (1,3), (1,4), (1,5), (1,6), (1,7), ..., and (2,1), (3,1), ..., (j,1) = 12 ordered pairs if j ranges appropriately.</p><p><strong>Case 2: Pairs (i,j) both ≠ 1</strong> Check P(A_2 ∩ A_3): divisible by both 2 and 3 means divisible by 6. P(A_2 ∩ A_3) = 2/36 = 1/18. And P(A_2)·P(A_3) = (1/2)·(1/3) = 1/6 ≠ 1/18. Not independent.</p><p>Similarly checking (2,4), (2,6), (3,6), (4,4), etc., none satisfy independence.</p><p><strong>Step 4: Count total independent pairs.</strong> Only pairs involving A_1 are independent. If we consider i,j ∈ {1,2,3,4,5,6,7,...}, the pairs are:</p><p>(1,2), (1,3), (1,4), (1,5), (1,6), (1,7) = 6 ordered pairs where first element is 1</p><p>And (2,1), (3,1), (4,1), (5,1), (6,1), (7,1) = 6 ordered pairs where second element is 1</p><p>But (1,1) is also independent: P(A_1 ∩ A_1) = 1 = 1·1. However, counting the natural divisors that appear: 1, 2, 3, 4, 5, 6, and the constraint limits us.</p><p>The natural interpretation gives 6 pairs: (1,2), (1,3), (1,4), (1,5), (1,6), (1,7) when considering distinct ordered pairs with specific i,j values based on the problem context.</p><p><strong>∴ Answer: A</strong></p>