<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] 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:
$var(X) = \sum P(X) X^2 - E[X]^2\\
2^2(\frac 19) + 1^2(\frac 49) - (2(\frac 19) + 1(\frac 49))^2\\
\frac {4}{9}$
$var(Y) = 2^2(\frac 1{36}) + 1^2(\frac {10}{36}) - (2(\frac 1{36}) + 1(\frac {10}{36}))^2\\
\frac {10}{36}$
$var (X+Y)= 2^2(\frac 1{4}) + 1^2(\frac {1}{2}) - (2(\frac 1{4}) + 1(\frac {1}{2}))^2\\
\frac {1}{2}$
$var (X+Y) = var(X) + var(Y) + 2cov(X,Y)\\
\frac {1}{2} = \frac {4}{9} + \frac {5}{36} + 2cov(X,Y)\\
cov(X,Y) = -\frac {1}{9}$
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