<p>A dice is rolled three times. Find the probability of getting a larger number than the previous number each time.</p>
Step-by-Step Solution
Key Concept: Count strictly increasing sequences (a < b < c) from three dice rolls where each outcome is from {1,2,3,4,5,6}. This equals C(6,3) divided by total outcomes 6³, since each 3-element subset has exactly one increasing arrangement.
<p><strong>Step 1:</strong> Identify what we need: Probability that 1st roll < 2nd roll < 3rd roll (strictly increasing)</p><p><strong>Step 2:</strong> Total possible outcomes when rolling a dice 3 times = 6³ = 216</p><p><strong>Step 3:</strong> Count favorable outcomes. We need to choose 3 distinct numbers from {1,2,3,4,5,6} and arrange them in increasing order. Each 3-element subset has exactly ONE strictly increasing arrangement.</p><p>Number of ways = C(6,3) = 6!/(3!×3!) = 20</p><p><strong>Step 4:</strong> These correspond to sequences like (1,2,3), (1,2,4), ..., (4,5,6)</p><p><strong>Step 5:</strong> Probability = 20/216 = 5/54</p><p>∴ Answer: A</p>
Correct Answer: A