<p>A dice is thrown six times, it being known that each time a different digit is shown. The probability that a sum of 12 will be obtained in the first three throws is</p>
<p>(1) \(\dfrac{5}{24}\)</p>
<p>(2) \(\dfrac{25}{216}\)</p>
<p>(3) \(\dfrac{3}{20}\)</p>
<p>(4) \(\dfrac{1}{12}\)</p>
Step-by-Step Solution
Key Concept: Since all six faces (1,2,3,4,5,6) must appear exactly once across six throws, the first three throws must show three distinct values from {1,2,3,4,5,6}. We need to count favorable outcomes (sum=12) and divide by total ways to arrange 3 distinct values in first three positions.
<p><strong>Step 1:</strong> Since each of the six throws shows a different digit, the six values {1,2,3,4,5,6} are distributed across the six throws, each appearing exactly once.</p><p><strong>Step 2:</strong> Find all ways the first three throws sum to 12 using three distinct numbers from {1,2,3,4,5,6}:</p><p>Possible sets: {1,5,6}, {2,4,6}, {2,5,5} ✗, {3,4,5}</p><p>Valid sets: {1,5,6}, {2,4,6}, {3,4,5}</p><p><strong>Step 3:</strong> Count arrangements for each set:</p><p>• {1,5,6}: 3! = 6 arrangements</p><p>• {2,4,6}: 3! = 6 arrangements</p><p>• {3,4,5}: 3! = 6 arrangements</p><p>Total favorable outcomes = 6 + 6 + 6 = 18</p><p><strong>Step 4:</strong> Total ways to choose and arrange 3 different numbers from {1,2,3,4,5,6} in first three positions = P(6,3) = 6 × 5 × 4 = 120</p><p><strong>Step 5:</strong> Probability = 18/120 = 3/20</p><p>∴ Answer: C</p>
Correct Answer: C