How to find PMF of max(X,Y) and (X|Y=1) The joint PMF of X and Y is:

I found the marginal PMF's:
$$P_X(x)=\begin{cases} 0.6 & , x = 0\\ 0.4 & , x  =1 & \\0 &,  \text{otherwise}
\end{cases}$$
$$P_Y(y)=\begin{cases} 0.3 & , y = -1\\ 0.2 & , y  =0 & \\ 0.5 &, y = 1\\0 &,  \text{otherwise}
\end{cases}$$
And now I want to find the PMF's of max(X,Y) and (X|Y=1). Do you have any ideas? Thanks!
 A: Do a simple reasoning....The rv "max" between $X,Y$ cannot take the value $-1$ as the minimum value of $X$ is zero...
Thus $Z=max(X,Y)$ can take only two values: $Z \in \{0;1\}$
What is the probability that $Z=0$? The only two possibilities are
$$P(Z=0)=P(X=0;Y=-1)+P(X=0;Y=0)=0.2$$
...as an immediate consequence
$$P(Z=1)=0.8$$
Are you able to attack the second question? It is not difficult...consider what will happen to $X$ when YOU KNOW that $Y=1$
A: (1). The outcomes of the random variable $Z=\max(X,Y)$ look like this
$$\begin{matrix}
\max(0,-1)=0 & \max(0,0)=0 & \max(0,1)=1 \\
\max(1,-1)=1 & \max(1,0)=1 & \max(1,1)=1 \\
\end{matrix}$$
So we have only two outcomes, $0$ and $1$.
$$P(\max(X,Y)=0)=P(X=0,Y=-1)+P(X=0,Y=0)=0.2$$
$$P(\max(X,Y)=1)=P(X=0,Y=1)+P(X=1,Y=-1)+P(X=1,Y=0)+P(X=1,Y=1)=0.8$$
(2). The conditional probability is found by computing
$$P(X=x|Y=1)=\frac{P(X=x,Y=1)}{P(Y=1)}$$
So
$$P(X=0|Y=1)=\frac{P(X=0,Y=1)}{P(Y=1)}=\frac{0.4}{0.5}$$
$$P(X=1|Y=1)=\frac{P(X=1,Y=1)}{P(Y=1)}=\frac{0.1}{0.5}$$
