The joint pdf of $X$ and $Y$ is given by:
$$f(x,y) = \frac{6}{7}\left(x^2+\frac{xy}{2}\right),\quad 0 < x < 1,\quad 0 < y < 2$$
Find $\;\displaystyle P \left( \left. X > \frac{1}{4}\; \right|\; X > Y\right).$
The method I attempted for solving this solution was using the conditional probability formula $$P(B | A) = \frac{P(B \cap A)}{P(A)}$$
The resulting problem I got was $$\frac{P(X > \frac{1}{4} \cap X > Y)}{P(X > Y)}$$
I converted this to $$\frac{\int_\frac{1}{4}^1 \int_0^x\frac{6}{7}(x^2+\frac{xy}{2})\,dy dx}{\int_0^1 \int_0^x\frac{6}{7}(x^2+\frac{xy}{2})\,dydx}$$
When I solved these integrals I got $$\frac{\frac{3825}{14336}}{\frac{15}{56}} = \frac{255}{256} = 0.99$$
This answer, specifically the top integral feels off so I was wondering if I went wrong anywhere or if the method I used is wrong entirely. Thanks!
You want to integrate over $\{1/4<x<1, 0<y<x\}$ and $\{0<x<1, 0<y<x\}$, as you have.
$$\mathsf P(X>1/4\mid X>Y)=\dfrac{\mathsf P(1/4<X<1,0<Y<X)}{\mathsf P(0<X<1,0<Y<X)}$$
Now $\int_0^x (x^2+xy/2)\mathsf d y=5x^3/4$, and $\int_z^1 x^3\mathsf d x=(1-z^4)/4$, so $$\mathsf P(X>1/4\mid X>Y)=\dfrac{(1-1/4^4)}{(1-0^4)}=\dfrac{255}{256}$$
So everything checks out okay.