<p>Let \(\omega\) be a complex cube root of unity with \(\omega \neq 1\). A fair die is thrown three times. If \(r_1\), \(r_2\) and \(r_3\) are the numbers obtained on the die, then the probability that \(\omega^{r_1} + \omega^{r_2} + \omega^{r_3} = 0\) is</p>
Step-by-Step Solution
Key Concept: Since ω is a primitive cube root of unity, ω³ = 1 and 1 + ω + ω² = 0. The sum ω^r₁ + ω^r₂ + ω^r₃ depends only on r₁ mod 3, r₂ mod 3, r₃ mod 3. We need the exponents (mod 3) to sum to 0 mod 3, which means we need either all three ≡ 0, or one each of 0, 1, 2 (mod 3).
<p><strong>Step 1:</strong> Recognize that ω³ = 1 and 1 + ω + ω² = 0 for primitive cube root of unity ω. Since ω^r only depends on r mod 3, classify die outcomes: {1,4} ≡ 1 (mod 3), {2,5} ≡ 2 (mod 3), {3,6} ≡ 0 (mod 3).</p><p><strong>Step 2:</strong> For ω^r₁ + ω^r₂ + ω^r₃ = 0, we need either:</p><p><strong>Case A:</strong> All three remainders ≡ 0 (mod 3): Choose from {3,6} three times → 2³ = 8 ways</p><p><strong>Case B:</strong> One remainder each of 1, 2, 0 (mod 3): </p><p>• Choose which position gets which remainder: 3! = 6 ways</p><p>• Each position: 2 choices → 2³ = 8 ways</p><p>• Total: 6 × 8 = 48 ways</p><p><strong>Step 3:</strong> Total favorable outcomes = 8 + 48 = 56</p><p>Total possible outcomes = 6³ = 216</p><p><strong>Step 4:</strong> Probability = 56/216 = 7/27</p><p>∴ Answer: C</p>
Correct Answer: C