Two disjoint random events You roll twice with four-sided die in which the numbers one and two occur with probability $\frac{1}{3}$, and the numbers three and four each with probability  $\frac{1}{6}$. Let X be the number of singles and Y the number of fours that occurred after two throws.
How do I create a table of probability function $p_{x,y}(x,y)=P\left \{X=x \wedge Y=y\right \}$?
This symbols at the end of this quations are little bit confusing to me.
$P(X=1)=\frac{1}{3}$, $P(X=2)=\frac{1}{3}$, $P(X=3)=\frac{1}{6}$ and $P(X=4)=\frac{1}{6}$. 
$P(X,Y)=P(x)P(Y)=\frac{1}{3}\frac{1}{6}=\frac{1}{18}$
So do I just write Table:
$x_i$  $P(X=x_i)$
1       $2*1/3$
4      $2*1/6$
Because there are two throws or?
How do I calculate $P\left \{X+Y>0\right \}$?
Do I just add them? 
$P\left \{X+Y>0\right \}=\frac{1}{3}+\frac{1}{6}$
 A: Start by noting that the question defines $X$ as the random variate representing the number of ones appearing in the outcomes of two throws.  So for example, 
$$
P(X=2) = \frac19\\ P(X=1) = \frac49 \\P(X=0) = \frac49
$$
Next you come to the point of the problem:  $P(X=1 \wedge Y=1)$ (for instance) is very different from $P(X=1) \cdot P(Y=1)$ because once you know that precisely one $1$ was rolled, that somewhat decreases the odds of precisely one $4$.  To compute the $X=1,Y=1$ entry in the table you would reason:
"To get one $1$ and one $4$ either the first roll must be $1$ and the second $4$ (probability $\frac13 \cdot \frac16 = \frac1{18}$ or the first must be $4$ and the second $1$ (same probability)." Since these two outcomes are mutually exclusive we can simply add the probabilities, giving $\frac19$ for that entry in the table."
The other entries are found in a similar way.
A: You cannot say $P(X,Y)=P(X)\cdot P(Y)$ as these events are not independent. 
Further, you need to account for different orderings. For example,
$$P(X=1, Y=1)=2\cdot\left(\frac{1}{3}\cdot\frac{1}{6}\right)=\frac{1}{9}$$
since we can get a one and then a four or a four and then a one.
Similarly
$$P(X=0, Y=1)=2\cdot\left(\frac{1}{2}\cdot\frac{1}{6}\right)=\frac{1}{6}$$
since we must get a two or three and then a four, or a four and then a two or a three.
You should find that the joint probability mass function is 
$$\begin{array}{|c|c|c|c|}
\hline
X/Y& 0 & 1 & 2 \\ \hline
 0& \frac{1}{4}& \frac{1}{6}&\frac{1}{36}\\ \hline
 1&  \frac{1}{3}& \frac{1}{9}&0\\ \hline
 2&  \frac{1}{9}& 0&0\\ \hline
\end{array}$$
As a check, the probabilities in this table do indeed sum to one, as required.
A: If X is the number of ones seen in two rolls.
Each roll has a $\frac 13$ chance of seeing a $1$
$P(X = 2) = (\frac {1}{3})^2\\
P(X = 0) = (1-\frac {1}{3})^2\\
P(X = 1)= 2(\frac 13)(\frac 23)$
If Y is the number of $4's$ seen in two rolls.
Each roll has a $\frac 16$ chance of seeing a $4$
$P(Y = 2) = (\frac {1}{6})^2\\
P(Y = 0) = (1-\frac {1}{6})^2\\
P(Y = 1)= 2(\frac 16)(\frac 56)$
Regarding $P(X+Y>0)$
Seeing a 4 and seeing a 1 on any roll are mutually exclusive events.   I think it is easier to look at the possibility that you roll a 4 or a 1.  The probability is $\frac 12$ on any given roll.
$P(X+Y>0) = 1-\frac 14$
Covariance:
$var(X) = \sum P(X) X^2 - E[X]^2\\
2^2(\frac 19) + 1^2(\frac 49) - (2(\frac 19) + 1(\frac 49))^2\\
\frac {4}{9}$
$var(Y) = 2^2(\frac 1{36}) + 1^2(\frac {10}{36}) - (2(\frac 1{36}) + 1(\frac {10}{36}))^2\\
\frac {10}{36}$
$var (X+Y)= 2^2(\frac 1{4}) + 1^2(\frac {1}{2}) - (2(\frac 1{4}) + 1(\frac {1}{2}))^2\\
\frac {1}{2}$
$var (X+Y) = var(X) + var(Y) + 2cov(X,Y)\\
\frac {1}{2} = \frac {4}{9} + \frac {5}{36} + 2cov(X,Y)\\
cov(X,Y) = -\frac {1}{9}$
