Let $f(x,y) = y^{-1}e^{-x/y}e^{-y}$ with $x,y >0$ and $0$ elsewhere.
Find $\mathbb{P}(\{X>1\} \mid \{Y = y\})$, $y >0$.
My attempt:
$$\mathbb{P}\{X>1, Y =y\} = \iint_{(1, +\infty) \times \{y\}} f(x,y) dxdy$$
And the integral gives 0 because I integrate over a line, so the probability is 0.
But I must have something wrong, but can't tell exactly what.
This is a conditional probability problem, and you had better figure out the conditional density $f_{X|Y}$ at first, which is defined as $$ f_{X|Y}(x,y)=\frac{f_{X,Y}(x,y)}{\int_{\mathbb{R}}f_{X,Y}(x,y){\rm d}x}. $$ Note that $$ \int_{\mathbb{R}}f_{X,Y}(x,y){\rm d}x=\int_{0}^{\infty}\frac{1}{y}e^{-x/y}e^{-y}{\rm d}x=-e^{-y}\int_{x=0}^{x=\infty}e^{-x/y}{\rm d}\left(-x/y\right)=e^{-y}. $$ Thus $$ f_{X|Y}(x,y)=\frac{1}{y}e^{-x/y}. $$ Thanks to this conditional density, $$ \mathbb{P}(X>1|Y=y)=\int_1^{\infty}f_{X|Y}(x,y){\rm d}x=\int_1^{\infty}\frac{1}{y}e^{-x/y}{\rm d}x=-\int_{x=1}^{x=\infty}e^{-x/y}{\rm d}\left(-x/y\right)=e^{-1/y}. $$