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 __________.

Solution

Maths sir · 3 steps

Do not count at least two hits directly: take away the zero-hit and one-hit cases from 1.

1. Set up binomial. Let X be the number of hits when n bombs are dropped. Each bomb hits independently with p = 1/2, so P(X = r) = C(n, r)(1/2)ⁿ.
2. Use the complement. The target is destroyed when X ≥ 2. So P(X ≥ 2) = 1 - P(X = 0) - P(X = 1) = 1 - (1 + n)/2ⁿ. We need this to be at least 0.99.

   1 - (n+1)/2ⁿ ≥ 99/100 2ⁿ ≥ 100 (n+1)

   n	2ⁿ	100(n+1)	2ⁿ ≥ 100(n+1)?
   9	512	1000	No
   10	1024	1100	No
   11	2048	1200	Yes

3. So. The condition first holds at n = 11, and 2ⁿ grows much faster than 100(n+1) after that. The minimum number of bombs is 11.

The common slip: 7. Requiring only one hit, so that 1/2ⁿ ≤ 0.01, gives 7, not 11. The one-hit case P(X = 1) must also be subtracted.

The idea to keep. For an at least k question with small k, subtract the few unwanted cases from 1. Then test whole numbers one by one, because an inequality like 2ⁿ ≥ 100(n+1) cannot be solved by algebra.

Answer from the NTA final answer key. Solution by PrepWiser.