Solving for number of trials for a given probability

Question

How to solve for the smallest $n$ such that $$1- 0.999^{n}-n\cdot0.001\cdot 0.999^{n-1}\geq 0.95$$

Answer 1

Consider the function $$f(n)=1-(1-a)^n-a n (1-a)^{n-1}$$ where $a$ is a small number. Use the binomial theorem or Taylor expansion to get $$f(n)=\frac{1}{2} a^2 n\left(n-1\right)+O\left(a^3\right)\approx \frac 12n^2a^2$$ So, a very first (bad !) approximation would be given by $$\frac 12 10^{-6} n^2=0.95 \implies n=100 \sqrt{190}\approx 1378$$ This gives a (poor) starting guess for Newton method.

I am sure that you can take it from here.