So we have a weapon with probability $\displaystyle p=0.95$ of working succesfully. We test $\displaystyle n$ weapons and the stockpile is replaced if the number of failures $\displaystyle X$ is at least one. How large must n be to have $\displaystyle P[X\geq 1]=0.99$?

a)Use exact binomial

$\displaystyle P[X\geq 1]=1-P[X=0]=0.99$ so $\displaystyle P[X=0]=0.01=$$\displaystyle \binom {n}{0}0.95^n*0.05^0=0.95^n \rightarrow n= \frac{log(0.01)}{log(0.95)}=90 $

b)Use normal approximation

I know we can use $\displaystyle Norm(np,np(1-p))$...

And then normalise this distribution to the standard normal form something like:

$\displaystyle P[X\geq 1]=1-P[X\leq 0.99] =1-\phi(\frac{(0.99-n0.95)}{\sqrt(n*0.95* 0.05)})=0.99$

If that is correct I have to solve that for n,.... how?

Also I doubt that rewriting $\displaystyle P[X\geq 1]=1-P[X\leq 0.99]$ is smart...