Given $f_x(x)=\frac{2x}{9}$ (for $0 < x < 3$)
and $f_{y|x}(y|x)=\frac{3y^2}{x^3}$ (for $0 < y < x$)
determine $P(Y<2)$
I already know that it is $\int_0^3 P(Y<2|X=x)f_x(x)dx$ but I dont know how to calculate the $P(Y<2|X=x)$ but... like im not sure what the limits are.
The joint pdf is given by $$f(x,y) = f_x(x) f_{y|x}(y|x)=\frac{2}{3}\frac{y^2}{x^2}$$ for $(0<x<3)$ and $(0<y<x)$ and zero otherwise.
Then, $$ P(Y<2) = \int_{-\infty}^{+\infty} dx \int_{-\infty}^2 dy \,f(x,y) $$
The limits are modified by removing the regions of $(x,y)$ where the density is zero: $$ P(Y<2) = \int_{0}^{3} dx \int_{0}^{min(x,2)} dy \,\frac{2}{3}\frac{y^2}{x^2} $$
The second set of limits can be treated by breaking the first integral into two pieces: $$ P(Y<2) = \int_{0}^{2} dx \int_{0}^x dy \,\frac{2}{3}\frac{y^2}{x^2} + \int_{2}^{3} dx \int_{0}^2 dy \,\frac{2}{3}\frac{y^2}{x^2} $$
This should be enough for you to get the answer...