<p>18. Two persons X and Y threw two dice each. The probability that Y throws a sum greater than X is when it is known that X throws a sum of at least 8, is ___.</p>
Step-by-Step Solution
Key Concept: Use conditional probability: P(Y > X | X ≥ 8). First find all outcomes where X ≥ 8, then count favorable outcomes where Y's sum exceeds each possible X value, weighted by their probabilities.
<p><strong>Step 1:</strong> Find P(X ≥ 8) by counting favorable outcomes when X throws two dice.</p><p>Sum = 8: (2,6), (3,5), (4,4), (5,3), (6,2) → 5 ways</p><p>Sum = 9: (3,6), (4,5), (5,4), (6,3) → 4 ways</p><p>Sum = 10: (4,6), (5,5), (6,4) → 3 ways</p><p>Sum = 11: (5,6), (6,5) → 2 ways</p><p>Sum = 12: (6,6) → 1 way</p><p>Total: 5 + 4 + 3 + 2 + 1 = 15 outcomes, so P(X ≥ 8) = 15/36</p><p><strong>Step 2:</strong> For each value of X, find P(Y > X) = favorable outcomes for Y / 36.</p><p>If X = 8: Y must be 9, 10, 11, or 12 → P(Y > 8) = 10/36</p><p>If X = 9: Y must be 10, 11, or 12 → P(Y > 9) = 6/36</p><p>If X = 10: Y must be 11 or 12 → P(Y > 10) = 3/36</p><p>If X = 11: Y must be 12 → P(Y > 11) = 1/36</p><p>If X = 12: No favorable outcomes → P(Y > 12) = 0</p><p><strong>Step 3:</strong> Apply conditional probability formula:</p><p>P(Y > X | X ≥ 8) = [5/36 × 10/36 + 4/36 × 6/36 + 3/36 × 3/36 + 2/36 × 1/36 + 1/36 × 0] / (15/36)</p><p>= [50 + 24 + 9 + 2 + 0] / (36² × 15/36)</p><p>= 85 / (36 × 15)</p><p>= 85 / 540 = <strong>17/108</strong></p>
Correct Answer: 17/108