So the joint p.d.f is f(x,y) = $c(x^2 + y^2)$ for $0\leq x\leq 1$ and $0\leq y \leq 1$
Now the steps I worked out is applying the formula $f(x|y) = \frac{f(x,y)}{f_{2}(y)}$
Which gave me f(x|y) = $\frac{3x^2+ 3y^2}{1+3y^2}$.
Assuming that is correct, the answer Pr(X$<\frac{1}{2}$|$Y=\frac{1}{2}$)should be $\frac{2}{7}$. I solved this to get that answer: $\int_{0}^{1/2} \frac{3x^2+ 3(\frac{1}{2})^2}{1+3(\frac{1}{2})^2}$ But the correct answer is $\frac{1}{3}$. I don't understand what I'm missing.