I have $N=100$ boxes with $K$ balls. For the experiment, I toss $K$ balls into the $N$ boxes, every ball is guaranteed to go into a box.
I want to know how big $K$ must be such that the probability of each box having a ball in it is at least $0.95$.
I was able to "solve" this by using Boole's inequality but that only provides me an approximation for low values of N.
Let $E_i$ be the event that the box $i$ is empty, and let $O_i=E_i^c$ be its complement (box is occupied). Notice first that the events $O_i$ are not independent.
But (de Morgan's Law) $$p=P(\cap O_i) = 1- P(\cup E_i) \tag{1}$$
and (Boole's inequality)
$$P(\cup E_i)\le \sum P(E_i)=n \left( \frac{n-1}{n}\right)^k \tag{2}$$
The latter comes from noting that the event $E_i$ correspond to having all $k$ balls in some of the others $n-1$ boxes. Each bll has a probability of $(n-1)/n$ of falling in another box. Then the probability that all balls fall in other boxes is the product over the $k$ balls, i.e $ P(E_i)=\left( \frac{n-1}{n}\right)^k$
Hence
$$ p\ge 1- n \left( \frac{n-1}{n}\right)^k \tag{3}$$
From this, setting $p=0.95$ and $n=100$ you can get $k$.
Because $n$ is large, we can (not necessary) approximate $(1- 1/n)^n\to e^{-1}$ to get $ k \approx n \log(20n)$ (as in robjohn's comment)
There are other ways of attacking this, but I guess that this is what you are after for. ( But, please, next time, try to add context, tell us what you know, what you have tried, etc. It's no fun to guess your context, and to imagine that (perhaps - I hope not) you will simply copy this answer for solving some homework without even thinking and trying to understand. The point is to learn. Welcome to MSE.)