Question
Let $X,Y$ be standard normal with $\operatorname{Cov}(X,Y)=\tfrac{1}{\sqrt{2}}$.
Compute $\mathbb{P}[X>0 \mid Y<0]$.
Solution

We write $$ \mathbb{P}[X>0 \mid Y<0] = \frac{\mathbb{P}(X>0,\, Y<0)}{\mathbb{P}(Y<0)} = 2,\mathbb{P}(X>0,\, Y<0), $$ since $\mathbb{P}(Y<0)=\tfrac12$.

Because $$ Y = \frac{1}{\sqrt{2}}\,X + \frac{1}{\sqrt{2}}\,Z, $$ with $Z$ independent of $X$, the vector $(X,Z)$ is rotationally symmetric in $\mathbb{R}^2$.

The event
$$ X>0,\qquad Y<0 $$ corresponds to the wedge where $X>0$ but $X+Z<0$.
Geometrically, the plane is divided into $8$ equal $45^\circ$ sectors, and exactly one of these sectors corresponds to $(X>0, X+Z<0)$. Thus $$ \mathbb{P}(X>0,\, Y<0)=\frac18. $$

Hence $$ \mathbb{P}[X>0 \mid Y<0] = 2 \times \frac18 = \frac14. $$