Find conditional probability given the joint probability

Question

Find the conditional distribution of $X$ and $Y$ given $Z = 0$ and the conditional distribution of $X$ and $Y$ given $Z = 1$

$P(X,Y|Z) = P(X, Y, Z) / P(Z) = P(X|Y,Z)P(Y|Z)$ So, For $Z = 0$:
$f(x=0,y=0|z = 0) = 0.405*0.45 = 0.18225$
$f(x=0,y=1|z = 0) = 0.045*0.05 = 0.00225$
$f(x=1,y=0|z = 0) = 0.045*0.45 = 0.02025$
$f(x=1,y=1|z = 0) = 0.005*0.05 = 0.00025$

For $Z = 1$:
$f(x=0,y=0|z = 1) = 0.125*0.25 = 0.03125$
$f(x=0,y=1|z = 1) = 0.125*0.25 = 0.03125$
$f(x=1,y=0|z = 1) = 0.125*0.25 = 0.03125$
$f(x=1,y=1|z = 1) = 0.125*0.25 = 0.03125$

Is the formula wrong? What is wrong with the computation?

Answer 1

The conditional distribution, summed on both $X,Y$ must be 1.

by definition,

$$P(X,Y|Z)=\frac{P(X,Y,Z)}{P(Z)}$$

the numerator is expressed in the matrix you showed.

$P(Z=0)=P(Z=1)=0.5$

Thus you conditional distribution is the same as you showed but with any joint probability multiplied by 2

$P(X=0,Y=0|Z=0)=0.81$

$P(X=0,Y=1|Z=0)=0.09$

$P(X=1,Y=1|Z=0)=0.09$

$P(X=1,Y=1|Z=0)=0.01$

total: $0.81+0.09+0.09+0.01=1$

nice end:)