Consider the joint PMF of discrete random variables shown below: $$\begin{array}{|l|l|l|l|} \hline & Y=0 & Y=1 & Y=2 \\ \hline X=0 & \frac{1}{6} & \frac{1}{6} & \frac{1}{8} \\ \hline X=1 & \frac{1}{8} & \frac{1}{6} & \frac{1}{4} \\ \hline \end{array}$$ Define $Z=E[X \mid Y]$, find PMF of $Z$
My try: It is known that $Z$ is random variable and is a function of $Y$. First i tried to find the conditional PMF $P(X=x \mid Y=y)$
For that we need marginal PMF of $Y$ which is: $$P(Y=y)= \begin{cases}\frac{7}{24}, & y=0 \\ \frac{8}{24}, & y=1 \\ \frac{9}{24}, & y=2\end{cases}$$ But i am stuck now?
As mentioned by others in comments,
$P(X=x|Y=y) = \dfrac{P((X =x) \cap (Y=y))}{P(Y=y)}$
$P(X = 1| Y = 0) = \dfrac 1 8 / \dfrac 7 {24} = \dfrac 3 7$
You already obtained, $P(Y = 0) = \dfrac{7}{24}$
$E(X| Y = 0) = 0 \cdot P(X = 0| Y = 0) + 1 \cdot P(X = 1| Y = 0)$
So, $Z = \dfrac{3}{7} ~ $ with probability $\dfrac{7}{24}$ or $ \displaystyle P\left(Z = \dfrac{3}{7}\right) = \frac{7}{24}$
Similarly find $Z = E(X| Y = 1)$ with probability $P(Y=1)$ and $Z = E(X| Y = 2)$ with probability $P(Y = 2)$