Probability
Classical Probability
Grade 12
Question:
<p>An unbiased dice, with faces numbered 1, 2, 3, 4, 5, 6, is thrown \(n\) times and the list of \(n\) numbers shown up is noted. Then find the probability that among the numbers 1, 2, 3, 4, 5, 6 only three numbers appear in this list and each number appears at least once.</p>
<p>\(\dfrac{(3^n - 3 \times 2^n + 3) \times {}^6C_3}{6^n}\)</p>
<p>\(\dfrac{3^n - 3 \times 2^n + 3}{6^n}\)</p>
<p>\(\dfrac{(3^n - 3 \times 2^n + 3) \times {}^6C_3}{3^n}\)</p>
<p>\(\dfrac{3^n - 2^n + 1}{6^n}\)</p>
Step-by-Step Solution
Key Concept: Use Stirling numbers of the second kind S(n,3) to count surjective functions from n throws onto exactly 3 distinct faces, then divide by total outcomes 6^n. The probability equals [3! × S(n,3)]/6^n, which simplifies to [S(n,3)]/6^n when accounting for selecting which 3 numbers appear.
<p><strong>Step 1:</strong> Total possible outcomes when throwing dice n times = 6<sup>n</sup></p><p><strong>Step 2:</strong> We need exactly 3 distinct numbers to appear, each at least once. First choose which 3 numbers from {1,2,3,4,5,6} appear: C(6,3) ways.</p><p><strong>Step 3:</strong> For 3 fixed numbers, count surjective functions from n throws onto these 3 numbers. This equals 3! × S(n,3), where S(n,3) is the Stirling number of the second kind (number of ways to partition n items into 3 non-empty subsets).</p><p><strong>Step 4:</strong> By inclusion-exclusion principle, surjections from n elements to 3 specific elements = 3<sup>n</sup> - C(3,1)×2<sup>n</sup> + C(3,2)×1<sup>n</sup> = 3<sup>n</sup> - 3×2<sup>n</sup> + 3</p><p><strong>Step 5:</strong> Total favorable outcomes = C(6,3) × (3<sup>n</sup> - 3×2<sup>n</sup> + 3) = 20(3<sup>n</sup> - 3×2<sup>n</sup> + 3)</p><p><strong>Step 6:</strong> Probability = [20(3<sup>n</sup> - 3×2<sup>n</sup> + 3)]/6<sup>n</sup> = [20(3<sup>n</sup> - 3×2<sup>n</sup> + 3)]/6<sup>n</sup></p><p>∴ Answer: A</p>
Correct Answer: A