Probability
Negative Binomial Distribution
Grade 12
Question:
<p>A fair die is thrown 20 times. The probability that on the 10th throw, the fourth six appears is</p>
<p>(1) \({}^{20}C_{10} \times 5^6/6^{20}\)</p>
<p>(2) \(120 \times 5^7/6^{10}\)</p>
<p>(3) \(84 \times 5^6/6^{10}\)</p>
<p>(4) none of these</p>
Step-by-Step Solution
Key Concept: The fourth six must appear on the 10th throw, meaning exactly 3 sixes in the first 9 throws and a six on the 10th. Use binomial probability combined with the requirement that the 10th position is fixed.
<p><strong>Step 1:</strong> For the fourth six to appear on the 10th throw:</p><ul><li>The 10th throw must show a six: P(six) = 1/6</li><li>Exactly 3 sixes must appear in the first 9 throws</li></ul><p><strong>Step 2:</strong> Probability of exactly 3 sixes in 9 throws:</p><p>P(3 sixes in 9 throws) = C(9,3) × (1/6)³ × (5/6)⁶</p><p><strong>Step 3:</strong> Calculate C(9,3):</p><p>C(9,3) = 9!/(3!×6!) = 84</p><p><strong>Step 4:</strong> Combined probability:</p><p>P = C(9,3) × (1/6)³ × (5/6)⁶ × (1/6)</p><p>P = 84 × (1/6)⁴ × (5/6)⁶</p><p>P = 84 × (5⁶)/(6¹⁰)</p><p>P = 84 × 15625/60466176</p><p>∴ Answer: C</p>
Correct Answer: C