<p>Two persons \(A\) and \(B\) throw two dice each. If \(A\) throws a sum of 9 then the probability of \(Y\) throwing a sum greater than that of \(X\) is</p>
<p>(a) \(\frac{2}{9}\)</p>
<p>(b) \(\frac{1}{9}\)</p>
<p>(c) \(\frac{1}{54}\)</p>
<p>(d) none of these</p>
Step-by-Step Solution
Key Concept: Once A fixes their outcome (sum = 9), B's probability is independent. We need P(B's sum > 9) given that B throws two dice with 36 equally likely outcomes.
<p><strong>Step 1:</strong> A has already thrown a sum of 9 (given condition). This is fixed and doesn't affect B's throw.</p><p><strong>Step 2:</strong> B throws two dice. Total possible outcomes = 36. We need P(B's sum > 9) = P(sum ∈ {10, 11, 12}).</p><p><strong>Step 3:</strong> Count outcomes for each sum:</p><ul><li>Sum = 10: (4,6), (5,5), (6,4) → 3 ways</li><li>Sum = 11: (5,6), (6,5) → 2 ways</li><li>Sum = 12: (6,6) → 1 way</li></ul><p><strong>Step 4:</strong> Total favorable outcomes = 3 + 2 + 1 = 6</p><p><strong>Step 5:</strong> Probability = 6/36 = 1/6</p><p>∴ Answer: A (probability = 1/6)</p>
Correct Answer: A