<p>The minimum number of times a fair coin needs to be tossed, so that the probability of getting at least two heads is at least 0.96, is ______. <b>(JEE Advanced 2015)</b></p>
Step-by-Step Solution
Key Concept: Use the complement: P(at least 2 heads) = 1 - P(0 or 1 heads). Find the minimum n where 1 - [C(n,0)(1/2)^n + C(n,1)(1/2)^n] ≥ 0.96, which simplifies to finding the smallest n where (n+2)/2^n ≤ 0.04.
<p><strong>Step 1:</strong> Set up using complement. P(at least 2 heads) = 1 - P(0 heads) - P(1 head)</p><p>P(0 heads) + P(1 head) = C(n,0)(1/2)^n + C(n,1)(1/2)^n = (1/2)^n + n(1/2)^n = (n+1)/2^n</p><p><strong>Step 2:</strong> We need: 1 - (n+1)/2^n ≥ 0.96</p><p>This gives: (n+1)/2^n ≤ 0.04</p><p><strong>Step 3:</strong> Test values systematically:</p><p>• n = 6: (7)/64 = 0.109 > 0.04 ✗</p><p>• n = 7: (8)/128 = 0.0625 > 0.04 ✗</p><p>• n = 8: (9)/256 = 0.0352 < 0.04 ✓</p><p><strong>Step 4:</strong> Verify for n = 8: P(at least 2 heads) = 1 - 9/256 = 247/256 ≈ 0.9648 ≥ 0.96 ✓</p><p>∴ Answer: 8</p>
Correct Answer: 8