<p>A die is thrown 4 times. Find the probability of getting at most two 6.</p>
Step-by-Step Solution
Key Concept: Use the complement approach: P(at most two 6s) = 1 - P(three or more 6s). This avoids calculating multiple cases and reduces computation to P(exactly three 6s) + P(exactly four 6s).
<p><strong>Step 1:</strong> Identify the setup. Each throw: P(getting 6) = 1/6, P(not getting 6) = 5/6. We need P(at most 2 sixes in 4 throws).</p><p><strong>Step 2:</strong> Use complement: P(at most 2) = 1 - P(3 or 4 sixes)</p><p><strong>Step 3:</strong> Calculate P(exactly 3 sixes): C(4,3) × (1/6)³ × (5/6)¹ = 4 × (1/216) × (5/6) = 20/1296</p><p><strong>Step 4:</strong> Calculate P(exactly 4 sixes): C(4,4) × (1/6)⁴ × (5/6)⁰ = 1 × (1/1296) × 1 = 1/1296</p><p><strong>Step 5:</strong> P(3 or 4 sixes) = 20/1296 + 1/1296 = 21/1296 = 7/432</p><p><strong>Step 6:</strong> P(at most 2) = 1 - 7/432 = 425/432 ≈ 0.984</p>
Correct Answer: 0.984