<p>In a precision bombing attack, there is a 50% chance that any one bomb will strike the target. Two direct hits are required to destroy the target completely. The number of bombs which should be dropped to give a 99% chance or better of completely destroying the target can be</p>
Step-by-Step Solution
Key Concept: Use the complement rule: P(at least 2 hits) ≥ 0.99 is equivalent to 1 - P(0 or 1 hit) ≥ 0.99, where each bomb independently has probability 0.5 of hitting. This reduces to finding the minimum n where P(X ≤ 1) ≤ 0.01 for binomial distribution with p = 0.5.
<p><strong>Step 1:</strong> Let X = number of bombs that strike the target out of n bombs dropped. X follows Binomial(n, 0.5).</p><p><strong>Step 2:</strong> We need P(X ≥ 2) ≥ 0.99, which means P(X ≤ 1) ≤ 0.01.</p><p><strong>Step 3:</strong> P(X ≤ 1) = P(X = 0) + P(X = 1) = (0.5)^n + n(0.5)^n = (0.5)^n(1 + n)</p><p><strong>Step 4:</strong> We need (0.5)^n(1 + n) ≤ 0.01, or equivalently (1 + n) ≤ 0.01 × 2^n</p><p><strong>Step 5:</strong> Testing values:</p><p>• n = 7: (1 + 7) = 8 and 0.01 × 2^7 = 0.01 × 128 = 1.28, so 8 > 1.28 ✗</p><p>• n = 8: (1 + 8) = 9 and 0.01 × 2^8 = 0.01 × 256 = 2.56, so 9 > 2.56 ✗</p><p>• n = 9: (1 + 9) = 10 and 0.01 × 2^9 = 0.01 × 512 = 5.12, so 10 > 5.12 ✗</p><p>• n = 10: (1 + 10) = 11 and 0.01 × 2^10 = 0.01 × 1024 = 10.24, so 11 > 10.24 ✗</p><p>• n = 11: (1 + 11) = 12 and 0.01 × 2^11 = 0.01 × 2048 = 20.48, so 12 < 20.48 ✓</p><p><strong>∴ Minimum number of bombs = 11</strong></p>
Correct Answer: B