<s>[INST] find conditional distribution $ \mathbb{P}\left(\frac{1}{4} \le Y \le \frac{3}{4} \bigm| X = \frac{1}{3}\right) $ $f(x,y) =x+y$,   $0<x<1$  ,$0<y<1$
find 
$ \mathbb{P}\left(\frac{1}{4} \le Y \le \frac{3}{4} \bigm| X = \frac{1}{3}\right)$
For now,
$$fx(x) = x+\frac{1}{2} $$,   $$fy(y) = y+\frac{1}{2} $$
and $$(y \mid x) = \frac{x + y}{x + 1/2}$$
i get the just plug in x=$\frac{1}{3}$
$$\frac{x + y}{1/3 + 1/2}$$
this is what i got so far,
however, how can i solve the Y? should i have to use integral? 
 [/INST] You are on your way. But please note that the conditional density of $Y$ given that $X=1/3$ is $\frac{1/3+y}{1/3+1/2}$, not $\frac{x+y}{1/3+1/2}$. Now integrate from $1/4$ to $3/4$.
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