If I draw 10 balls at random from a bag of 30 balls, what is the probability that I will have drawn all balls at least once after 10 turns? I thought of this question and I am unsure of how to solve it.
There are 30 balls in total in the bag. Each turn you remove 10 balls from a bag without replacement. Then you put them all back and take another turn. What is the probability that you will have drawn every ball at least once at the end of 10 turns?
I started with 0 overlap case (i.e. 2 turns, 20 in the bag), but that didn't give me much insight. Also, tried checking out the complimentary situation: $P($not seeing one ball$)$. I got 
$$
1 - \left ( \frac{29}{30} \times \ ...\ \times\frac{20}{21}\right )^{10} 
$$
But that does not seem right. Also doesn't account for which ball I chose to exclude.
 A: There are $30 \choose 10$ ways to select the balls at one turn and $29 \choose 10$ ways to select the balls skipping a specific ball.  We can do inclusion-exclusion.  We start with the ${30 \choose 10}^{10}$ ways to choose the balls and select the $30 \cdot {29 \choose 10}^{10}$ that are missing a ball, but we have subtracted the ones missing two balls twice and need to add them back.  Overall we get $${30 \choose 10}^{10}-30 \cdot {29 \choose 10}^{10}+{30 \choose 2}{28 \choose 10}^{10}-{30 \choose 3}{27 \choose 10}^{10}$$ ways to get all the balls.  I should keep going but the terms are probably shrinking fast.  The probability of getting all the balls is then $$\frac{{30 \choose 10}^{10}-30 \cdot {29 \choose 10}^{10}+{30 \choose 2}{28 \choose 10}^{10}-{30 \choose 3}{27 \choose 10}^{10}}{{30 \choose 10}^{10}} \approx 0.5773$$  Adding then next term in increased the value to $0.5831$, more of a change than I expected.  The correct answer is between these.
A: Suppose we ask  about the scenario where we  have $kn$ distinguishable
balls  with $k\le  n$ and  we draw  $n$ batches  of $n$  balls without
replacement and ask about the probability  that we have seen each ball
at  least  once.   This  yields  from  first   principles  with  $u_p$
representing a ball of type $p$ the generating function
$$\left([z^n] \prod_{p=1}^{kn} (1 + z u_p)\right)^n.$$
Now when we choose $q$ variables from  the $u_p$ and set them to zero,
setting  the  rest  to  one  then we  get  a  generating  function  of
configurations where the balls that  correspond to these variables are
missing, or possibly  some more. Hence we have  for the configurations
where none are missing by  inclusion-exclusion the closed form for the
probability
$${kn \choose n}^{-n} \sum_{q=0}^{kn} {kn\choose q} (-1)^q
\left([z^n] (1+z)^{kn-q}\right)^n$$
which yields
$$\bbox[5px,border:2px solid #00A000]{
{kn \choose n}^{-n}  \sum_{q=0}^{kn} {kn\choose q} (-1)^q
{kn-q\choose n}^n.}$$
In  the  present case  we  have  $k=3$  and  $n=10$ which  yields  the
probability
$${\frac {
36577493641520669177803175335325492081551459709262478876117528}{
63271363740585012990529845354903290101446377706425717867578125}}$$
which is approximately
$$0.5781050301.$$
