Calculating expectation and variance for having rolled 1 and 6 twice out of rolling a die 12 times First i have calculated the probability to get each possible number $\{1,2,3,4,5,6\}$ twice from $12$ rolls ($A$). We have:
$$Pr[A]=\frac{\binom{12}{2,2,2,2,2,2}}{6^{12}}.$$
Then there are 2 random variables:


*

*$X$-number of times that $1$ was received, and

*$Y$-number of times that $6$ was received.
Before calculating $E(X),Var(X),E(Y),Var(Y)$ i'm uncertain of how i should calculate the probabilities of X and Y 
 A: $Pr(X=x,Y=y)$ is the probability of getting $x$ throws equal to $1$ and $y$ throws equal to $6$.
Here, to calculate we can use the multinomial distribution, simplifying things to just refer to $1$'s, $6$'s, and "Others" where we don't bother breaking the "Others" category down further into each of the different numbers.
Applying the formula (which should be self-evident how it works) you get the probability is:
$$Pr(X=x,Y=y) = \binom{12}{x,y,(12-x-y)}\left(\frac{1}{6}\right)^x\left(\frac{1}{6}\right)^y\left(\frac{4}{6}\right)^{12-x-y}$$
So, the probability of getting exactly two $1$'s and exactly two $6$'s is going to be:
$$Pr(X=2,Y=2)=\frac{12!}{2!2!8!}\cdot \frac{4^8}{6^{12}}$$
As for calculating other metrics of the random variables $X,Y$ where $X$ and $Y$ represent the total number of $1$'s and $6$'s thrown respectively, @Vizag's hint is a good one, representing them each as sums of indicator random variables.
It should be clear that $Pr(X_i = 1) = \frac{1}{6}$, that $Pr(X_i,X_j) = \frac{1}{36}$, that $Pr(X_i,Y_i)=0$ and that $Pr(X_i,Y_j)=\frac{1}{36}$ for $i\neq j$.
Armed with this knowledge and remembering that expectation is linear so you can break apart an expectation of a sum as a sum of expectations, it should be clear how to calculate $E[X^2]$ and $E[XY]$ etc... to complete the later parts of the problem.
A: You can think of $X$ and $Y$ like this (which greatly simplifies things). 
Define an indicator variable $X_i$ such that $X_i = 1$ if we saw a $1$ on the $i^{th}$ throw and $0$ otherwise. Now convince yourself that, 
$$X = \sum_{i=1}^{12} X_i$$
Similarly define indicator variable $Y_i$ such that $Y_i = 1$ if we saw a $6$ on the $i^{th}$ throw and $0$ otherwise. 
Now note that: 
$$E[X] = \sum_{i=1}^{12} E[X_i], \quad \text{ By Linearity of expectations}$$
$$Var(X) = \sum_{i=1}^{12} Var(X_i), \quad \text { Because $X_i$'s are independent}$$
