I have the following problem with which I cannot solve. I have a very large population of birds e.g. 10 000. There are only 8 species of birds in this population. The size of each species is the same.
I would like to calculate how many birds I have to catch, to be sure in 80% that I caught one bird of each species.
On
If the population has $b$ birds divided equally between $s$ species then I think the probability that you have caught $k$ species after catching $n$ birds satisfies
$$P_{n+1,k}=P_{n,k-1}\frac{b(s-k+1)}{(b-n)s} +P_{n,k}\frac{bk - ns}{(b-n)s}$$
starting with $P_{0,0}=1$ and $P(0,k)=0$ for $k \gt 0$
It seems that with $b=10000$ and $s=8$, you have $P_{27,8} \approx 0.795$ and $P_{28,8} \approx 0.819$ making the answer $28$
As noted in a comment by @HJ_beginner, this is the Coupon Collector's Problem, with species of birds instead of coupons.
A First Course in Probability, Seventh Edition by Sheldon Ross gives the following solution to the Coupon Collector's Problem in Example 1e, section 4.1. Let's say there are $N$ types of coupons, and let $T$ be the number of the first draw in which you have a complete set of all $N$ coupons. Then $$P(T>n) = \sum_{i=1}^{N-1} \binom{N}{i} \left( \frac{N-i}{N} \right)^n (-1)^{i+1}$$ for $n>0$.
In your case, we have $N=8$ and we want the least $n$ such that $P(T > n) < 0.2$. By computation, the least such $n$ is $n=28$, with $P(T > 28) = 0.181464$.