How large must $n$ be to have $P[X \geq 1] = 0.99$ when $p = 0.95$?

Question

certain type of weapon has probability $p$ of working successfully. We test $n$ weapons, and the stockpile is replaced if the number of failures, $X$, is at least one. How large must $n$ be to have $P[X \geq 1] = 0.99$ when $p = 0.95$?

I only have one question about this exercise: is it advisable to use the central limit theorem to find $n$ or is it better to use other results, such as Chebyshov's inequality, for example? Any good suggestion is important.

Answer 1

(assuming independence)

EDIT: as note by @bof (thanks for correcting) I confused successes with failures.

Using the binomial, the question is to calculate

$$\mathbb{P}[X \geq 1]=1-\mathbb{P}[X =0]$$

Where X are the failures in a binomial $B(n;0.95)$

That is

$$1-\binom{n}{n}0.95^n\cdot 0.05^0=1-0.95^n$$

Now simply setting

$$1-0.95^n\geq 0.99$$

you get

$$n\geq \Bigg\lceil\frac{log0.01}{log0.95}\Bigg\rceil=90$$