The probability that the ratio of two independent standard normal variables is less than $1$ Let the independent random variables $X,Y\sim N(0,1)$. Prove that $P(X/Y < 1) = 3/4. $
Could anyone help me prove this analytically? Thanks. 
Progress: My first thought was to integrate the joint density function: $\dfrac{e^{-\frac{x^2}{2}-\frac{y^2}{2}}}{2\pi}$ but I'm not sure where to go from here. 

It is well-known that the  random variable $X/Y$ has the two-sided Cauchy density $\frac{1}{\pi(1+x^2)}$ for $-\infty<x<\infty$.  Thus 
$P(X/Y<1)=\int_{-\infty}^{1}\frac{1}{\pi(1+x^2)}dx=0.5+\int_{0}^{1}\frac{1}{\pi(1+x^2)}dx$ 
and so
$P(X/Y<1)=0.5+\frac{1}{\pi}\hbox{arctg}(1)=0.5+0.25=0.75$.
Note: A much simpler way is to consider the random vector $(X,Y)$ on the plane.  Then, 
$P(X/Y<1)=P(X/Y<1\mid Y>0)\times\frac12+P(X/Y<1\mid Y<0)\times\frac12\\
= (\frac12+\frac14) \times \frac12+ (\frac12+\frac14)\times\frac12=\frac34.$
