<p>Minimum number of times a fair coin must be tossed so that the probability of getting at least one head is more than 99% is __________.</p>
Step-by-Step Solution
Key Concept: Use the complement approach: P(at least one head) = 1 - P(all tails). Set this inequality greater than 0.99 and solve for n using logarithms.
<p><strong>Step 1:</strong> Use complement: P(at least one head) = 1 - P(all tails)</p><p><strong>Step 2:</strong> P(all tails in n tosses) = (1/2)^n</p><p><strong>Step 3:</strong> Set up inequality: 1 - (1/2)^n > 0.99</p><p><strong>Step 4:</strong> Rearrange: (1/2)^n < 0.01</p><p><strong>Step 5:</strong> Take logarithm: n·log(1/2) < log(0.01)</p><p><strong>Step 6:</strong> Simplify: n·(-log 2) < -2, so n > 2/log(2)</p><p><strong>Step 7:</strong> Calculate: n > 2/0.301 ≈ 6.64</p><p><strong>Step 8:</strong> Since n must be an integer and satisfy strict inequality, n ≥ 7</p><p>∴ Answer: 7</p>
Correct Answer: 7