If $$ f_x(x)= \begin{cases} \frac{x^2}9, & 0 \lt x \lt 3,\\ 0, & \text{otherwise},\end{cases} $$
$$ f_{y \mid x} = \begin{cases} \frac{4y^3}{x^4}, & 0 \lt y \lt x, \\ 0, & \text{otherwise}. \end{cases} $$ Find $P(Y \lt 1)$.
Working: I know that
(*)$ \ P(Y\lt 1) = \int_{0}^{3} P(Y \lt1 \mid X = x) \ f_x(x) \ dx $
and calculated that
$ P(Y \lt1 \mid X = x)= \int_{0}^{1}f_{y \mid x}(y \mid x) \; dy = \ \frac{1}{x^4}$,
but subbing that back into (*), I get a negative so it's obviously wrong, but i don't understand where I've went wrong! Any help is appreciated!
You were on the right track. The only problem with your solution is that you don't take in consideration that $P(Y<1∣X<1)=1$ since $0<y<x$. If you take that into consideration, you will find:
$$\ P(Y\lt 1) = \int_{0}^{3} P(Y \lt1 \mid X = x) \ f_x(x) \ dx$$ $$\ P(Y\lt 1) = \int_{0}^{1} 1* \ f_x(x) \ dx + \int_{1}^{3} P(Y \lt1 \mid X = x) \ f_x(x) \ dx$$ $$\ P(Y\lt 1) = \int_{0}^{1} \frac{x^2}{9} \ dx + \int_{1}^{3} \frac{1}{9x^2} \ dx$$