Conditional probability for bivariate random variables question: $$ f_{(X,Y)(x,y)}=
\begin{cases} 8xy, &0\le y\le x \le 1\\0.&\text{elsewhere} \end{cases} $$
Find $ P\left( X\le \frac{1}{2}|Y\ge \frac{1}{4}\right) $
What is wrong with my solution:
$$
\frac{\int_{\frac{1}{4}}^{\frac{1}{2}}\int_{\frac{1}{4}}^x8xydydx}{\int_{\frac{1}{4}}^{1}4y-4y^3dy}
$$
seeing that $ 4y-4y^3 $is the marginal pdf of Y

Regards @matt . If I may contribute. Isn't 
$$  P \left( \small{ X \le \frac{1}{2} | Y \ge \frac{1}{4} } \right) =  \frac{P(X \le \frac{1}{2} \cap Y \ge \frac{1}{4} )}{ P (Y \ge \frac{1}{4})}  \:\:\:\: ?$$
The numerator should be : $$ \int_{\frac{1}{4}}^{\frac{1}{2}} \int_{\frac{1}{4}}^{x} \: (8xy) \: dy \: dx $$
to avoid confusion, you could draw the $x$-$y$ diagram to make a clear view of the area. The area for the numerator is the area below the line $y(x) = x$, with $ \frac{1}{4} \le x \le \frac{1}{2} $, its a triangle.
Regards, Arief.
