Question

In a bombing attack, there is 50% chance that a bomb will hit the target. Atleast two independent hits are required to destroy the target completely. Then the minimum number of bombs, that must be dropped to ensure that there is at least 99% chance of completely destroying the target, is __________.

Hint:

Let X be the number of hits in n bombs, X ~ Binomial(n, 1/2). Require P(X \(\ge\) 2) \(\ge\) 0.99, i.e. \(\frac{n+1}{2^n} \le 0.01\).

Answer 1

Correct Answer: 11

Let n bombs be dropped; the number of hits X follows Binomial(n, 1/2).
P(X \(\ge\) 2) = 1 - P(X=0) - P(X=1) = 1 - \(\left(\frac{1}{2}\right)^n\) - \(n\left(\frac{1}{2}\right)^n\) = 1 - \(\frac{n+1}{2^n}\).
We need 1 - \(\frac{n+1}{2^n}\) \(\ge\) 0.99, i.e. \(\frac{n+1}{2^n} \le \frac{1}{100}\), i.e. \(2^n \ge 100(n+1)\).
n = 9: \(2^9\) = 512 < 1000, fails.
n = 10: \(2^{10}\) = 1024 < 1100, fails (probability of success = 1 - 11/1024 = 0.9893 < 0.99).
n = 11: \(2^{11}\) = 2048 \(\ge\) 1200, holds (probability = 1 - 12/2048 = 0.9941).
Minimum number of bombs = 11.
Answer: 11