<p>A coin is tossed \(n\) times. The probability of getting at least one head is at least \(\dfrac{99}{100}\). What is the minimum value of \(n\)?</p>
Step-by-Step Solution
Key Concept: Use the complement approach: P(at least one head) = 1 - P(no heads). Since P(no heads) = (1/2)^n, we need 1 - (1/2)^n ≥ 99/100, which means (1/2)^n ≤ 1/100.
<p><strong>Step 1:</strong> Use complement: P(at least one head) = 1 - P(all tails)</p><p>P(all tails) = (1/2)^n</p><p><strong>Step 2:</strong> Set up the inequality: 1 - (1/2)^n ≥ 99/100</p><p><strong>Step 3:</strong> Rearrange: (1/2)^n ≤ 1/100</p><p><strong>Step 4:</strong> Take logarithms: n·log(1/2) ≤ log(1/100)</p><p>n·(-log 2) ≤ -log 100</p><p>n·log 2 ≥ log 100 = 2 (since log₁₀100 = 2)</p><p><strong>Step 5:</strong> Solve for n: n ≥ 2/log₁₀(2) ≈ 2/0.301 ≈ 6.64</p><p><strong>Step 6:</strong> Verify: (1/2)⁷ = 1/128 < 1/100 ✓ and (1/2)⁶ = 1/64 > 1/100 ✗</p><p>∴ Minimum value of n = <strong>7</strong></p>
Correct Answer: 7