Probability of $(0,...,9)$ balls never being drawn on $10$ draws from $10$ balls with putbacks. I want to calculate the expected value and variance for the random variable
$$X = \text{number of balls which were never drawn}$$
when drawing from $10$t times from $10$ different balls with putbacks.
To calculate the expected value and variance I wanted to first find a closed formula for the probability distribution of $X$ on the numbers $(0-9)$. I tried to reduce it to a product of binomial distributions without success.
How can this problem be modeled? Is there another way to calculate the expected value or variance?
 A: If you use indicator random variables you can calculate expectation and variance without getting a pmf.
$$\text{Let } I_i = \begin{cases} 1: & \text{ball $i$ is never drawn} \\ 0: &\text{ball $i$ is drawn} \end{cases} \\ \mathsf E[I_i] = P(I_i = 1) = P(\text{ball $i$ is never drawn}) = .9^{10} \\ \mathsf E[X] = \mathsf E[I_0 + I_1 + \cdots + I_9] = 10\left(.9^{10}\right)$$
$$\mathrm{Var}(I_i) = P(I_i=1)(1-P(I_i=1)) = .9^{10}\left(1-.9^{10}\right) \\ \forall \ i \ne j, \ \mathrm{Cov}(I_i,I_j) = .8^{10}-(.9^{10})^2 \\ \mathrm{Var}(X) = \mathrm{Var}(I_0 + I_1 + \cdots I_9) = \sum_{i=0}^9\mathrm{Var}(I_i) + \sum_{i\ne j}\mathrm{Cov}(I_i,I_j) \\ =10(.9^{10})(1-.9^{10}) + 90(.8^{10}-.9^{20})$$
Reference: https://en.wikipedia.org/wiki/Indicator_function#Mean%2C_variance_and_covariance
A: If you are still interested in a closed form expression for the probability it is the following. Let $n$ be the number of balls, $m$ be the number of draws, and $k$ be the number of balls that were never drawn.
Then the probability in question is (credit to the comment of aschepler):
$$
p(k)=\frac1{n^m}{m \brace {n-k}}\frac{n!}{k!}
$$
where ${m \brace k}$ is the Stirling number of second kind.
For $n=m=10$ the probabilities read:
$$\begin{array}{r|r}
k& p\times10^{10}\\
\hline
0&3628800\\
1&163296000\\
2&1360800000\\
3&3556224000\\
4&3451442400\\
5&1285956000\\
6&171889200\\
7&6717600\\
8&45990\\
9&10\\
10&0
\end{array}$$
I would not however recommend to compute the expected value and variance using the above explicit expression for the probability. Instead resort to the recipe of the previous answer.
