<p>A pair of unbiased dice is rolled together till a sum of either 5 or 7 is obtained. The probability that 5 comes before 7 is</p>
Step-by-Step Solution
Key Concept: Model this as a Markov chain where we need P(get 5 before 7). Calculate P(5 on single roll), P(7 on single roll), and P(neither), then use the recursive probability equation: P(5 before 7) = P(5)/(P(5) + P(7)) when accounting for rolls that produce neither outcome.
<p><strong>Step 1:</strong> Find probabilities for a single roll of two dice.</p><p>P(sum = 5) = 4/36 = 1/9 [pairs: (1,4), (2,3), (3,2), (4,1)]</p><p>P(sum = 7) = 6/36 = 1/6 [pairs: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1)]</p><p>P(neither 5 nor 7) = 1 - 1/9 - 1/6 = 1 - 2/18 - 3/18 = 13/18</p><p><strong>Step 2:</strong> Set up the recursive equation. Let p = P(5 comes before 7).</p><p>On any roll: either we get 5 (probability 1/9), or we get 7 (probability 1/6), or we get neither (probability 13/18) and start over.</p><p>p = 1/9 + (13/18)·p</p><p><strong>Step 3:</strong> Solve for p.</p><p>p - (13/18)p = 1/9</p><p>(5/18)p = 1/9</p><p>p = (1/9) × (18/5) = 2/5</p><p><strong>Alternatively (Direct method):</strong> Among rolls that produce 5 or 7, the probability of getting 5 is:</p><p>P(5 | 5 or 7) = P(5)/(P(5) + P(7)) = (1/9)/(1/9 + 1/6) = (1/9)/(5/18) = 2/5</p><p>∴ Answer: A (2/5)</p>
Correct Answer: A