Let $X,Y$ be two i.i.d random variables with standard normal distribution which is the probability that the vector $(X,Y)$ to be in the second quadrant with distance to the origin greater than $2$?
What I think I am being asked is to calculate $$\mathbb{P}(X\leq 0,Y\geq 0, X^2+Y^2\leq 4)$$ How do I calculate that probability?
Let $R$ be the region corresponding to the set of points $(x,y)\in\mathbb{R^2}$ such that $x\leq 0$, $y\geq 0$ and $x^2+y^2>4$. Then by independence the joint density of $(X,Y)$ is given by $$ f(x,y)=\frac{1}{2\pi}\exp\left(-\frac{1}{2}(x^2+y^2)\right). $$ Then the required probability is given by $$ p=\int_R f(x,y)\, dx\, dy=\int_{\pi/2}^\pi\int_{2}^\infty\frac{1}{2\pi}\exp\left(-\frac{1}{2}r^2\right)r\, dr\,d\theta $$ by changing to polar coordinates. You should be able to compute the integral.
Let $\rho(x)$ be the usual standard normal density, $$ \rho(x)=\frac 1{\sqrt{2\pi}}\exp-\frac{x^2}2\ , $$ so we know the joint density for $(X,Y)$, it is $\rho(x)\rho(y)$ with respect to the standard Lebesgue mass $dx\; dy$, so we have to calculate: $$ \begin{aligned} p &= \mathbb{P}(X\leq 0,Y\geq 0, X^2+Y^2\color{red}{\geq} 2^2) \\ &= \int_{\substack{(x,y)\in\Bbb R^2\\x<0\\y>0\\x^2+y^2\ge 2^2}} \rho(x)\; \rho(y)\; dx\; dy \\ &\qquad\text{ change of variables }(x,y)=(r\cos t,r\sin t) \\ &= \int_{r\ge 2} \int_{t\in(\pi/2,\pi)} \frac 1{2\pi} \exp-\frac {r^2}2 \cdot r\; dr\; dt \\ &= \frac \pi 2 \int_{r\ge 2} \frac 1{2\pi} \exp-\frac {r^2}2 \cdot \frac 12d(r^2) \\ &= \frac 14\left[\ -\exp-\frac{r^2}2\ \right]_{r=2}^\infty =\frac 14\exp(-2)\approx 0.0338338208091532\dots\ \ . \end{aligned} $$
In such cases i usually also start a computer simulation:
sage: import numpy
sage: N = 10**8 # samples
sage: X = numpy.random.normal(size=N)
sage: Y = numpy.random.normal(size=N)
sage: V = (X^2+Y^2)[ (X<0) & (Y>0) ]
sage: nr_goodCases = len( V[ V>= 4 ] )
sage: nr_goodCases
3383058
sage: print "Proportion of good cases ~ %.8f" % (nr_goodCases / N)
Proportion of good cases ~ 0.03383058
this time...
