Let $X$ and $Y$ have the joint pdf $f(x,y)=8x(1-y)$, $0<y<1$, $0<x<1-y$. Compute $P(Y<X\mid X \leqq \frac{1}{4})$.
I know:
$$P(A\mid B) = \frac{P(A \cap B)}{P(B)}$$
Therefore I first compute $P(B \leqq \frac{1}{4})=\int_0^{\frac{1}{4}}4x$ to satisfy the denominator. To calculate the numerator I need to find the intersection of $A$ and $B$:
$$\int_0^{\frac{1}{4}} \int_0^x 8x(1-y)\:\mathrm{d}y\:\mathrm{d}x$$
Am I on the right track? My answer is extremely close, but not exactly correct.
You have the right integral for $P(A\cap B)$ but the wrong expression for $P(B)$, which should be $$ P(B) = \int_{x=0}^{1/4} dx \int_{y=0}^{1-x} 8x(1-y) dy$$ The upper limit in the $y$ integral comes from the fact that $$ x < 1-y \Leftrightarrow y < 1-x$$