Let X and Y have the joint pdf $f(x,y) = 8x(1-y)$, $0 < y < 1$, and $0 < x < 1-y$
Compute $P(Y < X |X\leq1/4)$
I worked out that the marginal pdf of $x$ is $4x$ and that the conditional pdf is $2(1-y)$ I'm just not sure where to go from there.
Answer = 29/93
$P(Y<X|X\leq \frac{1}{4})=\frac{P(Y<X\leq \frac{1}{4})}{P(X \leq \frac{1}{4})}$. Calculate $P(Y<X\leq \frac{1}{4})=\int_0^{1/4}\int_0^{x}8x(1-y)dydx=\frac{29}{768}$ and compute $P(X\leq \frac{1}{4})=\frac{1}{8}$. The final answer comes to be $\frac{29}{96}$. Please recheck the numerical computations.