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The question is if probability of bomb hitting a target is 50% and 2 direct hits will destroy the target completely. How many bombs must be dropped to give a 99% chance or better of destroying the target completely? I assume this question can be done using negative binomial distribution where favourable number of results is 2? Am I right this is an example in my book and author used binomial distribution. Please suggest if any other possible method is applicable? Thanks

Onix
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1 Answers1

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If you drop $n$ bombs, you do not destroy the target iff all the bombs miss or just one bomb hits the target. That happens with probability $\frac{1}{2^n}+\frac{n}{2^n}$, hence you are looking for the smallest $n$ that grants $$ \frac{n+1}{2^n}\leq \frac{1}{100}.$$ Such $n$ is $\color{red}{11}$.

Jack D'Aurizio
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  • Sir, you give the exact answer what the book says. But what is the intuition behind your solution or you used the same binomial distribution.. – Onix Nov 15 '15 at 16:27
  • Sorry sir I get the method but how do you calculate the probability to be 1/2^n +n/2^n – Onix Nov 15 '15 at 16:29
  • How do you compute the probability that all the $n$ bombs miss? How do you compute the probability that all the bombs but one miss? – Jack D'Aurizio Nov 15 '15 at 16:42
  • By using formula for binomial distribution for this ques.i.e.nC0(1/2^n)& nC1(1/2^n) Thanks – Onix Nov 15 '15 at 16:56
  • @AbhishekSharma: exactly. $\binom{n}{0}=1,\binom{n}{1}=n$. – Jack D'Aurizio Nov 15 '15 at 16:59