<p>A pair of dice is rolled together till a sum of either 5 or 7 is obtained. Find the probability that 5 comes before 7.</p>
Step-by-Step Solution
Key Concept: Use the concept of conditional probability and geometric series: in each roll, calculate P(sum=5), P(sum=7), and P(neither), then find the probability that 5 appears before 7 by summing over all possible rounds where 5 first occurs.
<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 equation for probability that 5 comes before 7.</p><p>Let P = probability that 5 comes before 7</p><p>P = P(5 on first roll) + P(neither on first roll) × P(5 comes before 7)</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>P(1 - 13/18) = 1/9</p><p>P(5/18) = 1/9</p><p>P = (1/9) × (18/5) = 2/5</p><p>∴ Answer: <strong>2/5</strong></p>
Correct Answer: 2/5