Probability of hitting an aircraft is .001 for each shot. How many shots should be fired so that the probability of hitting with two or more shots os above 0.95? From my textbook, i know how to calculate the number of trials if the question was to calculate the probability of hitting with atleast one attempt. Using binomial PDF, solving for 1-P(X=0)>0.95 gets me the answer in that case. Don't know what to do when P(X=1) comes into play

  • $\begingroup$ Rather than $1-P(X=0)=P(X\geq 1)$., instead use $1-P(X=0)-P(X=1)=P(X\geq 2)$. As for how to calculate $P(X=0)$ and $P(X=1)$., that should be made clear from the definition of a binomial distribution. $\endgroup$ – JMoravitz Jun 6 '18 at 3:26

It is much the same process. $~\mathsf P(X\geq 2)=1-\mathsf P(X\leq 1)$

Hence what you now want is the smallest trial amount $n$ for which $X\sim\mathcal{Bin}(n, 0.001)$ gives: $$1-\mathsf P(X{=}0)-\mathsf P(X{=}1)\geq 0.95$$

Now you know $\mathsf P(X=k)=\tbinom{n}k\cdot 0.001^k\cdot0.999^k$ for all $k\in\{0 .. n\}$.

So $\mathsf P(X{=}0)=0.999^{n}$ and $\mathsf P(X=1)=n \cdot 0.001\cdot0.999^{n-1}$ .

Thus just find the smallest $n$ such that $$1- 0.999^{n}-n\cdot0.001\cdot 0.999^{n-1}\geq 0.95$$

Hint: $0.999^n=0.999\cdot0.999^{n-1}$

  • $\begingroup$ Thank you but I had already reached this point and I am having a hard time solving this equation $\endgroup$ – Sahin Jun 6 '18 at 14:16

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