Action...

Solution 1.

It is sufficient to find the maximum n such that n bombs are required to be dropped to give a less than 1% chance of not destroying the target, i.e. either no bomb or only one bomb strike the target. Thus

1/2^n+n/2^n < 0.01,

(n+1)/2^n < 0.01.

If n = 10, then (n+1)/2^n = 11/1024 > 11/1100 = 0.01.

If n = 11, then (n+1)/2^n = 12/2048 < 20/2048 < 0.01.

Hence the maximum number of bomb drops is n = 11.

Solution 2.

Let A _k be the event that k bombs are required to strike the target twice, i.e. one and only one out of the first k-1 and the kth bomb drops strike the target. Thus Pr(A _k) = (k-1) _C _1 *(1/2)^(k-1) *(1/2) = (k-1)/2^k. The question asks to find the minimum n such that P = sum _(k = 2)^(n) {Pr(A _k)} >= 0.99. Now P = sum _(k = 2)^(n) {Pr(A _k)}, P = 1/2^2+2/2^3+3/2^4+...+(n-2)/2^(n-1)+(n-1)/2^n. Multiplying both sides by 1/2, we get

P/2 = 1/2^3+2/2^4+3/2^5+...+(n-2)/2^n+(n-1)/2^(n+1). Subtracting this from the previous equation, we get

P/2 = 1/2^2+1/2^3+1/2^4+...+1/2^n-(n-1)/2^(n+1). Multiplying both sides by 1/2, we get

P/4 = 1/2^3+1/2^4+1/2^5+...+1/2^(n+1)-(n-1)/2^(n+2). Subtracting this from the previous equation, we get

P/4 = 1/2^2-n/2^(n+1)+(n-1)/2^(n+2),

P = 1-n/2^(n-1)+(n-1)/2^n. Now we seek the minimum n such that P >= 0.99,

1-n/2^(n-1)+(n-1)/2^n >= 0.99, or

n/2^(n-1)-(n-1)/2^n <= 0.01.

If n = 10, then n/2^(n-1)-(n-1)/2^n = 10/512-9/1024 = 11/1024 > 11/1100 = 0.01.

If n = 11, then n/2^(n-1)-(n-1)/2^n = 11/1024-10/2048 = 11/2048 < 11/1100 = 0.01. Hence the minimum number of bomb drops is n = 11.

Solution 3 (by Nishan Ranabhat).

The target is destroyed if at least two out of n(say) bombs dropped hits it directly. But 3 or more bombs may hit the target so;

[nC2 + nC3 +....+nCn] (1/2)^n >= 0.99

which give (2^n - n-1)/2^n >=0.99 => 2^n = 100(n+1) => n = 11 ( by trial method).