Conditional probability given that $X$ and $Y$ are uniform Suppose that $X$ and $Y$ are uniformly distributed on $0\lt |x| + |y| \lt 1$. How do I find $P(Y\gt 1/4 ~\vert ~X=1/2)?$
 A: The intuition says that the answer is $1/4$. Computation is really unnecessary, but for the sake of more complicated situations let us do it.  
We are told, I think, that the pair $(X,Y)$ is uniformly distributed on the square with corners $(1,0)$, $(0,1)$, $(-1,0)$, and $(0,-1)$.  This square has area $2$, so our joint density function $f_{X,Y}(x,y)$ is $1/2$ inside the square and $0$ elsewhere. 
First we find the marginal density $f_X(x)$.  This is 
$$\int_y f_{X,Y}(x,y)\,dy.$$
In our case (for positive $0\le x< 1$), we are integrating from $y=x-1$ to $y=1-x$, and we get $1-x$. (More precisely, $f_X(x)=1-|x|$ on the interval $|x|<1$). 
Next we find the conditional density of $Y$ given $X=x$, that is, $f_Y(y|X=x)$. To do this, divide the joint density by the marginal density. Take in particular $x=1/2$.  Since $f_X(1/2)=1/2$, we have
$$f_Y(y|X=1/2)=1$$
for $-1/2<y<1/2$.  
Finally, for our answer, "integrate" the constant function $1$ from $y=1/4$ to $y=1/2$. We get $1/4$.  (The process would be more interesting with a problem in which the answer is not intuitively clear.) 
A: Ben Derrett's comment allows conditioning here so agreeing with yoyo, 
if $Y$ is uniformly distributed such that  $\frac{1}{2}+|Y| \lt 1$, 
then $Y$ is uniformly distributed on $(-\frac{1}{2},\frac{1}{2})$, 
and so $\Pr(Y\gt \frac{1}{4}|X=\frac{1}{2}) = \frac{1}{4}$.
