<p>A pair of four dice is thrown independently three times. The probability of getting a score of exactly 9 twice is</p>
Step-by-Step Solution
Key Concept: Identify this as a binomial probability problem where each trial (throwing two dice) has a fixed probability of getting sum = 9, and we need exactly 2 successes in 3 independent trials.
<p><strong>Step 1: Find probability of sum = 9 in one trial (two dice)</strong></p><p>Favorable outcomes: (3,6), (4,5), (5,4), (6,3) → 4 outcomes</p><p>Total outcomes: 36</p><p>p = 4/36 = 1/9</p><p><strong>Step 2: Find probability of NOT getting sum 9</strong></p><p>q = 1 - 1/9 = 8/9</p><p><strong>Step 3: Apply binomial probability formula</strong></p><p>P(X = 2) = C(3,2) × p² × q¹</p><p>P(X = 2) = 3 × (1/9)² × (8/9)</p><p>P(X = 2) = 3 × (1/81) × (8/9)</p><p>P(X = 2) = 3 × 8/729</p><p>P(X = 2) = 24/729 = 8/243</p><p><strong>∴ Answer: C (8/243)</strong></p>
Correct Answer: C