Let $p=0.1$ be the probability that a produced dice is broken. I now want to calculate how many dices have to be produced to assure with at least $99$% probability that at least 1000 dices are not broken.
This obviously follows a binomial distribution: $X \sim Bin(n, 0.9)$, where $n$ is the variable I want to determine. I know that for large $n$ we can approximate by a normal distribution.
$$\mathbb{P}(X \geq 1000) \geq 0.99 => 1 - \mathbb{P}(X \leq 1000 - 1) \geq 0.99 => 1 - \Phi(\frac{999-0.9n}{\sqrt{0.9n(1-0.9)}}) \geq 0.99$$
How can I determine $n$?
$$1-\Phi\left(\frac{1000-0.9n}{\sqrt{0.1n(1-0.1)}}\right)\geq 0.99$$ is equivalent to $$ \Phi\left(\frac{1000-0.9n}{\sqrt{0.1n(1-0.1)}}\right)\leq 0.01 $$ Use table of standard normal distribution or Excel function Norm.S.Inv to get $$ \frac{1000-0.9n}{\sqrt{0.1n(1-0.1)}}\leq \Phi^{-1}(0.01)\approx-2.326 $$ And then solve quadratic inequality with respect to $\sqrt{n}$ $$ 1000-0.9n \leq -2.326 \sqrt{0.09n}=-2.326\times0.3\sqrt{n} $$ The result is $n\geq 1137.26$. Since $n$ is integer, we should take $n\geq 1138$.
We can also check the accuracy by Excel function Binom.Dist: $$=1-\text{BINOM.DIST}(999;1138;0.9;1) = 0.991376330173266 > 0.99,$$ and $$=1-\text{BINOM.DIST}(999;1137;0.9;1) = 0.989229871716899 < 0.99.$$