<s>[INST] 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? 
$P\left \{X+Y>0\right \}=\frac{1}{3}+\frac{1}{6}$
 [/INST] 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.
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