I'm trying to solve a problem with various steps. There's the following joint pdf
$f_{XY}(x,y)= \begin{cases} x+y, &\text{if }0<x<1 \text{ and } 0<y<1\\0, &\text{otherwise} \end{cases} $
and one of the questions states that you need to find $P(Y<0.6|X>0.9)$. Given that I need the conditional density function, first I got the marginal distribution of $X$ which is $f_X(x)=x+\dfrac 12$, and then the conditional distribution would be $f_{Y|X=x}(y)=\dfrac{x+y}{x+\dfrac 12}=\dfrac {2(x+y)}{2x+1}$.
My question is the following: in order to obtain $P(Y<0.6|X>0.9)$, what do I need to do? Since X is not equal but rather larger than $0.9$, do I calculate the conditional distribution with the result of integrating $f_X$ from $0.9$ to $1$, and then divide $(x+y)$ by that result? All examples I've found X asume X takes a fixed value.
Directly use the definition for conditional probability : $$\mathsf P(Y<0.6\mid X>0.9)~{=\dfrac{\mathsf P(Y<0.6\cap X>0.9)}{\mathsf P(X>0.9)}\\=\dfrac{\int_0^{0.6}\int_{0.9}^1 (x+y)~\mathsf d x~\mathsf d y}{\int_0^1\int_{0.9}^1 (x+y)~\mathsf d x~\mathsf d y}}$$
Then evaluate those two integrals and simplify the resulting ratio.