variance of a sum We draws 10 from a finite population of 30, without replacement! 15 members out of 30 have the number 1, 10 members have the number 2 and 5 members have the number 3.
The Expected value is 
$$\mathbb E(X)= 1 \cdot \frac12+2 \cdot \frac13+3 \cdot \frac 16 = \frac 53 $$
Variance
$$V(X)= \left( 1 -\frac53 \right)^2 \cdot \frac12+ \left( 2-\frac53 \right)^2 \cdot \frac13+ \left( 3-\frac53 \right)^2 \cdot \frac16 = \frac59 $$
I need the Variance of $$X_{1}+...+X_{10}$$ and $$Var( \frac1{10}( X_{1}+...+X_{10}))$$
I found this formula $$V(X_{1}+...+X_{10})= \sum\limits_{i=1}^{10} \operatorname{Var}(X_i)+\sum\limits_{i\neq j}^{} \operatorname{Cov}(X_i,X_j)$$
But how i can solve this? 
With Wolframalpha?
thx

It is simpler to work it out directly. Since the total of the sample $Z = X_1+\ldots+X_{10}$ does depend on the order of sampling of each individual components, the distribution of $Z$ would be the same if, instead of sampling without replacement, we sampled 10 elements at once.
Suppose a sample contains $k_1$ balls with number 1, $k_2$ balls with number 2, and $k_3$ balls with number 3. The probability of obtaining such a configuration is:
$$
  p(k_1, k_2, k_3) = \mathbb{P}(K_1=k1,K_2=k_2,K_3=k_3) = \frac{ \binom{15}{k_1} \binom{10}{k_2} \binom{5}{k_3}}{ \binom{30}{10} } \mathsf{1}_{k_1+k_2+k_3=10}
$$
The triple $(K_1,K_2,K_3)$, thus follows the multi-variate hypergeometric distribution.
The mean is
$$
  \mathbb{E}(Z) = \mathbb{E}(K_1 + 2K_2+3K_3) = \mathbb{E}(K_1) + 2 \mathbb{E}(K_2) + 3 \mathbb{E}(K_3) = 1 \cdot \frac{10}{30} 15 + 2 \cdot \frac{10}{30} 10 + 3 \cdot \frac{10}{30} 5 = \frac{50}{3}
$$
The variance then
$$ \begin{eqnarray}
   \mathbb{Var}(Z) &=& \mathbb{Var}(K_1) + 4 \mathbb{Var}(K_2) + 9 \mathbb{Var}(K_3) + 4 \mathbb{Cov}(K_1,K_2) + 6 \mathbb{Cov}(K_1,K_3) + 12 \mathbb{Cov}(K_2,K_3)   \\
    &=& \frac{50}{29} + 4 \cdot \frac{400}{261} + 9 \cdot \frac{250}{261} + 4 \cdot \left( -\frac{100}{87}\right) + 6 \cdot \left( - \frac{50}{87} \right) + 12 \cdot \left( -\frac{100}{261} \right) = \frac{1000}{261}
 \end{eqnarray}
$$

