Probability Theory - Sum of Probabilities I am given that $Y$ and $Z$ are discrete random variables and $W$ and $X$ are continuous with $Y,Z\in \{0,1\}$. I am asked to find $P(Y=1)$ and I have used

$$P(Y=1)=\sum_zP(Z=z,Y=1)$$

but I'm not sure if this is correct as I think that I may have to sum over the other variables. Some help would be great!
Here are the probabilities I am given:

$$P(W=w\mid X=x,Y=y,Z=z)=\frac{1}{\sqrt{2\pi}}e^{-\frac{(w-z)^2}{2}}$$
$$P(X=x\mid Y=y,Z=z)=\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}$$
$$P(Y=0\mid Z=0)=0.1$$
$$P(Y=1\mid Z=0)=0.9$$
$$P(Y=0\mid Z=1)=0.3$$
$$P(Y=1\mid Z=1)=0.7$$
$$P(Z=0)=0.2$$
$$P(Z=1)=0.8$$

 A: Basic Approach.  Use the fact that
$$
P(Y = 1) = P(Y = 1, Z = 0) + P(Y = 1, Z = 1)
$$
combined with the general rule that
$$
P(A, B) = P(A \mid B) P(B)
$$
You do not need the conditional distributions of $W$ and $X$; they are superfluous for this (part of the) problem.
A: If $Z$ is a random variable that is equal to $z_1$ with probability $1/3$ and to $z_2$ with probability $1/12$ and to $z_3$ with probability $7/12$, then, since $1/3+1/12 + 7/12=1$, there are no further possibilities.  It is true that there may be other random variables, thus:
$$
\begin{array}{|c|c|c|c|c|}
& \hline Z = z_1 & Z=z_2 & Z=z_3 & {}\!\!\!\!{} & \text{total} \\
\hline
U=1 & 1/4 & 1/12 & 1/2 & {}\!\!\!\!{} & 5/6 \\
\hline
U=2 & 1/12 & 0 & 1/12 & {}\!\!\!\!{} & 1/6 \\
\hline \text{total} & 1/3 & 1/12 & 7/12 & {}\!\!\!\!{} & 1 \vphantom{\frac {\displaystyle\sum} 1} \\
\hline 
\end{array}
$$
However, you see that all six cells in this table are still accounted for by the three events that $Z=z_1$, $Z=z_2$, and $Z=z_3$.
Hence
\begin{align}
\Pr(Y=1) & = \Pr\Big( (Y=1\ \&\ Z=z_1) \text{ or } (Y=1\ \&\ Z=z_2) \text{ or } (Y=1\ \&\ Z=z_3) \Big) \\[10pt]
& = \Pr(Y=1\ \&\ Z-z_1) + \Pr(Y=1\ \&\ Z=z_2) + \Pr(Y=1\ \&\ Z=z_3).
\end{align}
So summing over all possible values of $Z$ is enough.
