Probability Brownian motion is up at 1 and down at 2

Let $B$ be a standard Brownian motion. Compute $\mathbb{P}[B_1>0\text{ and }B_2<0]$.

$1/8$. Write $B_1=Z_1$ and $B_2=Z_1+Z_2$ with $Z_1,Z_2$ independent standard normals. The event $Z_1>0,\;Z_2<-Z_1$ is a $45^\circ$ wedge of the plane, and by rotational symmetry its probability is $45/360=1/8$.