<p><strong>260.</strong> The probability of occurrence of a multiple of 2 on one dice and a multiple of 3 on the other dice if both are thrown together, is:</p>
<p>(a) \(\dfrac{7}{36}\)</p>
<p>(b) \(\dfrac{1}{3}\)</p>
<p>(c) \(\dfrac{1}{6}\)</p>
<p>(d) \(\dfrac{11}{36}\)</p>
Step-by-Step Solution
Key Concept: Identify favorable outcomes where one die shows a multiple of 2 (2,4,6) AND the other shows a multiple of 3 (3,6), accounting for both orderings and their intersection.
<p><strong>Step 1:</strong> Identify favorable outcomes for each condition.</p><p>Multiples of 2 on a die: {2, 4, 6} → 3 outcomes</p><p>Multiples of 3 on a die: {3, 6} → 2 outcomes</p><p><strong>Step 2:</strong> Count favorable cases for "multiple of 2 on one AND multiple of 3 on the other."</p><p>Case 1: Multiple of 2 on first die, multiple of 3 on second die</p><p>Outcomes: (2,3), (2,6), (4,3), (4,6), (6,3), (6,6) → 3 × 2 = 6 outcomes</p><p>Case 2: Multiple of 3 on first die, multiple of 2 on second die</p><p>Outcomes: (3,2), (3,4), (3,6), (6,2), (6,4), (6,6) → 2 × 3 = 6 outcomes</p><p><strong>Step 3:</strong> Avoid double-counting (6,6) is counted in both cases.</p><p>Total favorable outcomes = 6 + 6 - 1 = 11</p><p><strong>Step 4:</strong> Calculate probability.</p><p>Total possible outcomes = 36</p><p>Probability = 11/36</p><p>∴ Answer: D (11/36)</p>
Correct Answer: D