How to compute $\mathbb{P}(3x > -y > x \land x > 0 \land y < 0)$? Knowing that both $x \sim \mathcal{N}(0,1)$ and $y \sim \mathcal{N}(0,1)$ ($x,y$ independent from each other), I want to compute
$$\mathbb{P}(3x > -y > x \land x > 0 \land y < 0)$$
I'm aware that if two events are independent, I can take the two probabilities and multiply. E.g., if I were only interested in 
$$ \mathbb{P}(x > 0 \land y < 0) $$
I would simply compute $\mathbb{P}(x > 0) \mathbb{P}(y < 0)$.
However, in the problem described above, the condition $3x > -y > x$ is clearly not independent from the two other conditions $x >0 \land y <0$. I think that I should re-formulate the condition to $3x > -y \land -y > x$ ... but that still leaves the issue that the conditions are somewhat dependent from each other.
How can I approach such a problem?
 A: Since $X$ and $Y$ are independent Gaussians, they are also jointly normaly distributed. So, one can just take the pdf of a bivariate standard normal distribution and adjust the integrals according to the given conditions. With $Z = -Y$ and $f(x,z)$ denoting the bivariate normal distribution, I thought about the problem like this:
$ \quad P(3x > -y > x \land x > 0 \land y < 0) \\
=P(3x > z > x)  ; \ \ x,z>0  \\
=P(3x > z) - P(x > z); \ \ x,z>0   \\
=P(3x - z > 0) - P(x - z > 0); \ \ x,z>0   \\
= \int_0^{\infty} \int_0^{3x} f(x,z)dzdx - \int_0^{\infty} \int_0^{x} f(x,z)dzdx.$
Since the analytical solution at this point wasn't easy, I have to admit that I just used WolframAlpha. As a result I got: 
$\quad \int_0^{\infty} \int_0^{3x} f(x,z)dzdx - \int_0^{\infty} \int_0^{x} f(x,z)dzdx \\
= \frac{tan^{-1}(3)}{2\pi} - \frac{1}{8} \approx 0.073792. $
A: $\mathbb{P}(3x>-y>x \ \& \ x>0 \ \& \ y<0) = \mathbb{P}(3x>-y>x >0)$
Maybe this will help you on start.
Then, let's divide by x ( that is greater than 0)
$ \mathbb{P}(3x>-y>x >0) = \mathbb{P}(3>-\frac{y}{x}>1, x>0)$$
