Sampling with replacement with multiple attempts Let us consider sampling over $N$ objects in a bowl. $Q$ sampling attempts are made, each attempt consisting in picking $K_i\leq N$ objects, with $i = 1,\dots, Q$. After each attempt, the objects are put back in the bowl.
How to calculate the probability that, after the $Q$ attempts, each object of the bowl has been picked at least once?
For example, let us assume $N$=4, $Q$=2, $K_1$=2, $K_2$=3. Let us denote the objects as (1,2,3,4). In the first attempt, there are ${4}\choose{2}$ = 6 possibilities, i.e. ((1,2) (1,3) (1,4), (2,3), (2,4), (3,4)). In the second attempt, there are ${4}\choose{3}$ = 4 possibilities, i.e. ((1,2,3), (1,2,4), (1,3,4), (2,3,4)). The set of all possible combinations is therefore given by $6 \cdot 4$=24 combinations. From inspection, I can see that 12 combinations out of the 24, are those who cover all the 4 elements. Therefore result is 12/24=1/2. But how to come up with a general formula??
 A: This answer doesn't answer the general question. It only covers the special case Q=2 that OP asked for in a comment.
When $Q=2$ we have two attempts $K1$ and $K2$ where $1<=K1<=N$ and $1<=K2<=N$. Further, $K1+K2>=N$.
In the first attempt $K1$ unique numbers will be picked.
In the second attempt there will be $N-K1$ unique numbers and $K1$ duplicates (of the first attempt) to pick from. Further it's know that $K2 >= N-K1$.
So for the whole experiment to be a succes, we must pick all $N-K1$ unique numbers in the $K2$ picks. The number of combinations giving succes is then found as:
$$NumberOfCombinationGivingSucces = \binom{K2}{N-K1}$$
Now consider the case where $K2>=K1$
In this case the number of duplicates picked in the second attempt will be in the range "NumbersToPick - NumberOfUniqueNumbers" to "NumbersOfDuplicates". Or as a formula the range is from $K2-(N-K1)$ to $K1$. So we can calculate the total number of combinations like:
$$TotalNumberOfCombinations = \sum_{i=K2-(N-K1)}^{K1}{\binom{K2}{i}}$$
So with the restrictions a) $Q=2$ b) $K1+K2>=N$ and c) $K2>=K1$ the probability of picking all N numbers in the two attempts $K1$ and $K2$ is:
$$P = \frac{\binom{K2}{N-K1}}{\sum_{i=K2-(N-K1)}^{K1}{\binom{K2}{i}}}$$
Using the same principle the case $K2<K1$ can be calculated.
