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The probability of a bomb hitting a bridge is $1/2$. Two direct hits are needed to destroy it. What is the least number of bombs required so that the probability of the bridge being destroyed is greater than $0.9$?

Selecting two bombs $\binom{n}{2}$

hence probability is $\frac{1}{2}\binom{n}{2}$ which should be greater than 0.9 from this I got $n=6$. but answer is $7$

gt6989b
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Icandoahandstand99
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  • check the accepted answer here https://math.stackexchange.com/questions/747929/binomial-distribution-finding-the-number-of-trials-given-probability-and-succes – adhg Apr 24 '17 at 19:12

2 Answers2

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The probability that a bomb will not hit the bridge is $1-\frac12=\frac12$.

The probability that of $n$ bombs, none hits the bridge is $\left(\frac12\right)^n$.

The probability that of $n$ bombs, exactly $1$ hits the bridge is $n\frac12\left(\frac12\right)^{n-1}=n\left(\frac12\right)^n$.

So, the probability that the bridge is not destroyed is $1 - \left(\left(\frac12\right)^n+n\left(\frac12\right)^n\right) = 1 - (n+1)\left(\frac12\right)^n$.

$1 - (n+1)\left(\frac12\right)^n > 0.9$ gives $(n+1)\left(\frac12\right)^n < 0.1$.

You can check that $7$ is the smallest positive integer that satisfies this inequality.

wythagoras
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  1. Probability of all bombs missing the bridge is?
  2. Probability of exactly 1 bomb hitting the bridge is?
  3. Probability that neither (1) nor (2) happens is $p_n = ?$

Now, as you correctly mention, you need so find the least integer solution to $p_n > 0.9$, which is indeed 7.

gt6989b
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