You will randomly select 10 balls from the box with replacement what is $E(\bar X)$ A box contains 100 numbered balls - 21 with the number 1, 36 with the number 2 and 43 with the number 3.  You will randomly select 10 balls from the box with replacement and you take the mean of the sum of the 10 balls
What is $E(\bar{ X})$ and $Var(\bar {X})$
I know that there are 21 different averages ranging from 1 to 3 in increments of 0.1. There are a total of 66 combinations, but there must be a trick to do this without having to write out a 2 page long equation to calculate $E(\bar X)$ and an even longer $Var(\bar X)$
 A: Denote $X_1, X_2, X_3$ be the number of times that the three balls with number 1, 2, and 3 respectively, are selected in the 10 random draws. Since with replacement, $X=(X_1, X_2, X_3)$ follows a multinomial distribution with probabilities $p_1 = 0.21, p_2 = 0.36, p_3 = 0.43$. Following the distribution, we have
$$\mathbf{E}(X_i) = np_i; \hspace{20pt} \mathbf{Var}(X_i) = np_i(1-p_i); \hspace{20pt} \mathbf{Cov}(X_i, X_j) = -np_ip_j ( i \neq j).$$
We then have $$\bar{X} = \frac{1*X_1 + 2*X_2 + 3*X_3} {n} = \frac{X_1 + 2X_2 + 3X_3}{10} $$
$$ \mathbf{E}(\bar{X}) = \frac{1}{10} \{ \mathbf{E}(X_1) + \mathbf{E}(2X_2) + \mathbf{E}(3X_3) \}=p_1 + 2p_2 + 3p_3$$ $$=0.21 + 2*0.36 + 3*0.43 = 2.22\\$$
Similarly, $$\mathbf{Var}(\bar{X}) = \frac{1}{100} \{\mathbf{Var}(X_1) + 4\mathbf{Var}(X_2) + 9\mathbf{Var}(X_3) +4\mathbf{Vov}(X_1, X_2) + 12\mathbf{Cov}(X_2, X_3) + 6\mathbf{Cov}(X_1, X_3)\}.$$
EDIT:
The $X_1, X_2, X_3$ are correlated, so there should be covariance parts. I just added those.
Note that $\mathbf{Var}(aX + bY) = a^2\mathbf{Var}(X) + b^2\mathbf{Var}(Y) + 2ab\mathbf{Cov}(X,Y).$
END EDIT.
You can do the simple calculation by yourself.
Hope it helps. 
