Variance of summation of dependnent Bernoulli variables

Question

Put 10 identical balls into 3 distinct boxes randomly.
Define two random variables as follows.
$1\le i\le3$, $0\le k\le10$
$X_i=1$ if i-th box continas k balls.
$X_i=0$ otherwise.
$Y_k$=the number of boxes containing k balls.
$Y_k=\sum_{i=1}^3 X_i$
Want to find: Expectation and Variance of $Y_k$.

For each $i$, $X_i$ follows Bernoulli distribution, $E(X_i)=p={10 \choose k}(\frac{1}{3})^k(\frac{2}{3})^{10-k}$
Hence $E(Y_k)=\sum_{i=1}^3 E(X_i)=3E(X_i)={10 \choose k}(\frac{1}{3})^{k-1}(\frac{2}{3})^{10-k}$.
I could find expectation of $Y_k$ but I'm now strugling to find variance of $Y_k$.
If $k$ is sufficiently large, $X_i$'s are not mutually independent.
Clearly, $Var(Y_k) \neq \sum_{i=1}^3 Var(X_i)$.
So I tried to find Moment-generating function of $Y_k$ but I failed since $E(e^{tY_k})= E(e^{tX_1})E(e^{tX_2})E(e^{tX_3})$ may not hold without Independence.
Nevertheless, it seems that finding mgf of $Y_k$ is the only way to get $Var(Y_k)$.

Answer 1

Finding the MGF is not the only way to get the variance in question (and finding the MGF in this case seems quite hard) Here is one way:

For the rest of this question, fix k.

What you are trying to find is $E(Y^2)$ (then you can find the variance since you know the mean), which is $E(X_1 + X_2 + X_3)^2 = E[X_1 ^2 + X_2 ^ 2 + X_3 ^2] + 2E[X_1 X_2] + 2E[X_1 X_3] + 2E[X_2 X_3]$. Note that $X_i^2 = X_i$ since $X_i$ takes values 0 or 1.

Now, what is $X_i X_j$? It is 1 if and only if both of them are 1. Thus, you should compute the probability that both boxes have exactly k balls in them.