Computing $P(Y > 3X \mid Y > 0)$ where $X$ and $Y$ are i.i.d. standard normal How can I compute $P(Y > 3X \mid Y > 0)$ where $X$ and $Y$ are i.i.d. standard normal? The solution that I have is pretty unclear to me:
"The key is that $N(0, 1)^2$ is cyclically symmetric. When plotting the distributions, the p.d.f. will be cyclically symmetric about the origin. Then, one can perform a geometric probability calculation to obtain an answer in terms of $\arctan$".
I'm pretty confused, and I'm wondering if someone can please explain the solution. I tried using Bayes' Rule, which led me nowhere. I don't quite see how to visualize $N(0, 1)^2$, or even how it's related to the problem.
I just drew this picture in Desmos for reference:

 A: Hint: draw (preferably on a piece of paper) a sketch of a circle centered at the origin. Mark (by shading) the region corresponding to the event $(Y > 0)$.  Then, mark (by cross-hatching) the sub-region corresponding to the event $(Y >0)\cap Y > 3X)$. What fraction of the region $(Y > 0)$ is the region $(Y >0)\cap Y > 3X)$?
A: The key point is about the circular symmetry in this particular case, a property of using a standard multivariate normal distribution which has the joint density $\frac1{2\pi}e^{-(x^2+y^2)/2}$, so all points of equal radius from the origin have the same $x^2+y^2$ and so the same density.
The graph below shows a sample of $10000$ points and it illustrates this.  You are then trying to find the probability that a particular point is coloured red given that it is red or pink, and the circular symmetry means this is just a ratio of angles

R code for this
set.seed(2020)
X <- rnorm(10000)
Y <- rnorm(10000)
rpbg <- ifelse(Y > 0, ifelse(Y > 3*X, "red", "pink"),
                      ifelse(Y > 3*X, "black", "grey")) 
plot(X, Y, col=rpbg)

