a problem with joint density Let $X,Y,Z$ be three random variables that are determined as :
$$X:\Omega \longrightarrow \{1,2\}$$
$$Y:\Omega \longrightarrow \{1,2,3\}$$
$$Z:\Omega \longrightarrow \{1,2\}$$ 
(1)  I have to determine the value of $\Delta$ such that $$f_{X,Y, Z}(x,y,z)= \frac{1}{\Delta}*x*y*z$$ can be their density function.
$ \sum_{x=1}^2 \sum_{y=1}^3 \sum_{z=1}^2 f_{X,Y, Z}(x,y,z)= \frac{1}{\Delta}*(1+2)*(1+2+3)*(1+2)=1  \Rightarrow  \Delta=54$
(2)  Determine $P(x=1,y  \ge 2, z=1)$
(3)  Determine $P(x=2,y +z=4)$
(4) Determine the marginal density of (x,y) e the marginal density of x
But I'm not sure about the other points.

2) The event $\{Y\geq 2\}$ is the union of two disjoint events $\{Y=2\}$ and $\{Y=3\}$. Therefore
$$P(X=1, Y\geq 2, Z=1) = P(X=1, Y=2, Z=1)+P(X=1, Y=3, Z=1).$$
Use your probability mass function $$f_{X,Y, Z}(x,y,z)= \frac{xyz}{54}$$ to calculate both probabilities.
3) $Y+Z=4$ an two cases: $Y=2,Z=2$ or $Y=3, Z=1$. 
4) 
$$P(X=x, Y=y)=\sum_{z=1}^2 f_{X,Y,Z}(x,y,z).$$
$$P(X=x)=\sum_{y=1}^3 \sum_{z=1}^2 f_{X,Y,Z}(x,y,z).$$
