Product of three independent random variables We are given three independent random variables $X, Y, Z$.
$X$ has the following Bernoulli distribution: $P(X=1)=\frac{3}{4}, \ P(X=0) = \frac{1}{4}$
$Y$ has a uniform distribution on the interval $[-1,1]$
$Z$ has a uniform distribution on the interval $[0,2]$
How to establish the distribution of  random variable $W=XY - (1-X)Z$?
I know how to solve a similar problem, that is if we are given a random variable $X$ and its distribution and are asked to find the distriobution of $g(X)$, for example $X^2, \ aX+b, $ etc
But I don't know what to do if there are several random variables and of different types of distributions.
Could you help?
 A: Using the tower rule and the independence of our random variables we write that
$F_W(x)=P(W(X,Y,Z)<x)=E[P(W(X,Y,Z)<x|X)]=$
$P(W(X,Y,Z)<x|X=1)\frac{3}{4}+P(W(X,Y,Z)<x|X=0)\frac{1}{4}=$
$P(W(1,Y,Z)<x)\frac{3}{4}+P(W(0,Y,Z)<x)\frac{1}{4}.$
Since $W(1,Y,Z)=Y,\text{ and }W(0,Y,Z)=-Z$ we have finally $$F_W(x)=P(Y<x)\frac{3}{4}+P(-Z<x)\frac{1}{4}.$$
Given that $Y$ is uniformly distributed over $[-1,1]$ and $-Z$ is uniformly distributed over $[-2,0]$, 
$$P(Y<x)=\begin{cases}0 &\text{ if } x<-1\\
\frac{1}{2}(x+1) &\text{ if }  -1\le x \le 1\\
1&\text{ if } x>1
\end{cases},$$
and 
$$P(-Z<x)=\begin{cases}0 &\text{ if } x<-2\\\frac{1}{2}x+1& \text{ if }  -2\le x \le 0\\
1&\text{ if } x>0
\end{cases}.$$ 
The rest is easier to be drawn than told. Here is $F_W(x)$ the distribution function of $W$:

Edited to better explain the first steps of the calculation
Rather than referring to the tower rule we can do the first steps directly:
$P(W(X,Y,Z)<x)=$
$P(\{W(X,Y,Z)<x\}\cap \{X=1\}\cup \{W(X,Y,Z)<x\}\cap \{X=0\})=$
$P(\{W(X,Y,Z)<x\}\cap \{X=1\})+P(\{W(X,Y,Z)<x\}\cap \{X=0\})=$
$P(\{W(1,Y,Z)<x\}\cap \{X=1\})+P(\{W(0,Y,Z)<x\}\cap \{X=0\}).$
And now, here comes independence:
$P(\{W(1,Y,Z)<x\}\cap \{X=1\})=P(W(1,Y,Z)<x)P(X=1)=P(W(1,Y,Z)<x)\frac{3}{4}$
$P(\{W(0,Y,Z)<x\}\cap \{X=0\})=P(W(0,Y,Z)<x)P(X=0)=P(W(0,Y,Z)<x)\frac{1}{4}.$
