What is $P(X > 0 \mid X + Y > 0)$ given that $X, Y$ are i.i.d standard normal?
Solution 1:
The answer is obviously $3/4$ without needing any calculation.
The reasoning is that the joint density is radially symmetric, so that the following probabilities are all equal: $$\Pr[0 < Y < X] = \Pr[0 < X < Y] = \Pr[X < Y < 0] = \Pr[Y < X < 0]\\ = \Pr[0 < Y < -X] = \Pr[0 < -X < Y] = \Pr[-X < Y < 0] = \Pr[Y < -X < -0].$$ In exactly $3$ of these cases, $X > 0$ and $X+Y > 0$; in half of these cases $X+Y > 0$. So the conditional probability is $3/4$.