Random Variable X and Y has a joint probability density function.
$$f_{X, Y} (x, y) =\begin{cases} cxy& 0 \leq x \leq y, 0 \leq y \leq5\\ 0 & \text{otherwise }\\ \end{cases} $$
(a) Find $f_{Y | X}(y | x)$
(b) $P(Y \leq 4 | X = 3)$
$(a)$
$$f_X(x) = \int_{0}^{5} cxy dy = cx/2(y^2)\bigg|_{0}^{5} = \frac{cx25}{2}$$ with support $0 \leq x \leq y$
$$f_{Y | X}(y | x) = \frac{cxy}{cx25/2} = \frac{2y}{25}$$
$(b)$
$$f_{Y, X}(y | X=3) = \frac{2y}{25}$$
$$\int_{0}^{4} f_{Y, X}(y | X=3) dy = \int_{0}^{4} \frac{2y}{25} dy = \frac{2}{50}\left(y^2 \right)\bigg|_{0}^{4} = \frac{16}{25}$$
Apparently, this is wrong, why so?
The pdf only has support for $y>x,$ so actually $$ f_X(x) = \int_x^5cxy\;dy.$$