<p>A pair of fair 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: Use recursive probability: the event occurs immediately (with probability of getting 5) or occurs later (if neither 5 nor 7 is rolled).
<p><strong>Step 1:</strong> When rolling two dice, $P(\text{sum} = 5) = \frac{4}{36} = \frac{1}{9}$ (outcomes: (1,4), (2,3), (3,2), (4,1)).</p><p><strong>Step 2:</strong> $P(\text{sum} = 7) = \frac{6}{36} = \frac{1}{6}$ (outcomes: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1)).</p><p><strong>Step 3:</strong> $P(\text{neither 5 nor 7}) = 1 - \frac{1}{9} - \frac{1}{6} = 1 - \frac{2}{18} - \frac{3}{18} = \frac{13}{18}$.</p><p><strong>Step 4:</strong> Let $p$ be the probability that 5 comes before 7. Then: $p = \frac{1}{9} + \frac{13}{18} \cdot p$.</p><p><strong>Step 5:</strong> $p - \frac{13}{18}p = \frac{1}{9}$, so $\frac{5}{18}p = \frac{1}{9}$, giving $p = \frac{2}{5} = 0.4$.</p>
Correct Answer: C