<p>The minimum number of times one has to toss a fair coin so that the probability of observing at least one head is at least 90% is __________.</p>
Step-by-Step Solution
Key Concept: The probability of at least one head equals 1 minus the probability of all tails. Set up the inequality 1 - (1/2)^n ≥ 0.9 and solve for the minimum integer n.
<p><strong>Step 1:</strong> Use the complement rule. P(at least one head) = 1 - P(all tails)</p><p><strong>Step 2:</strong> Set up the inequality: 1 - (1/2)^n ≥ 0.9</p><p><strong>Step 3:</strong> Simplify: (1/2)^n ≤ 0.1, which gives (1/2)^n ≤ 1/10</p><p><strong>Step 4:</strong> Taking logarithms: n·log(1/2) ≤ log(1/10), so -n·log(2) ≤ -log(10)</p><p><strong>Step 5:</strong> Therefore: n ≥ log(10)/log(2) = 1/log₁₀(2) ≈ 3.32</p><p><strong>Step 6:</strong> Verify: n = 3: 1 - (1/8) = 7/8 = 0.875 < 0.9 ✗</p><p>n = 4: 1 - (1/16) = 15/16 = 0.9375 > 0.9 ✓</p><p>∴ Answer: 4</p>
Correct Answer: 4