Use Points in Euclidean space to find probability based on distributions? Given a string at the origin in  Euclidean real space to a random point (x,y) where x~N(2,1) and y~N(0,1), how can one determine the probability that the string will need to be longer > 3 units from origin?
Introductory work:
I recognize to use distance formula: square root($x^2$ + $y^2$) > $3^2$, and to square both sides.
I recognize that squaring vector x will produce a non-central $x^2$ and squaring vector y will produce a central $x^2$. 
I don't know what to do to solve from here. I have read previous problems on Euclidean distance but do not need to solve for the function. I need to somehow find the probability answer.
 A: Solution:


*

*If |y| is bigger than 3, the string is always longer than 3 units,
regardless of the value of x. The chance that the standard normal
distribution will be less than -3 or greater than +3 is $\text{erfc}\left(\frac{3}{\sqrt{2}}\right)$ which is about 0.0026998

*For -3 < y < 3, the chance that y = y0 (for any y0) is the PDF
of the standard normal at y0 or $\frac{e^{-\frac{\text{y0}^2}{2}}}{\sqrt{2 \pi }}$

*If y=y0, we want to compute the chance that x^2 + y0^2 > 3^2, or
x^2 > 3^2 - y0^2 or |x| > Sqrt[3^2-y0^2].

*The chance that x < -Sqrt[3^2-y0^2] is just the CDF of x's
distribution to the value, which is $\frac{1}{2} \text{erfc}\left(\frac{\sqrt{9-\text{y0}^2}+2}{\sqrt{2}}\right)$

*The chance that x > Sqrt[3^2-y0^2] is 1 minus the CDF of x's
distribution to the value, which is: $1-\frac{1}{2}\text{erfc}\left(-\frac{\sqrt{9-\text{y0}^2}-2}{\sqrt{2}}\right)$

*Since the two events above don't overlap, the total chance that x
will be large enough to make the total length bigger than 3 is the
sum of the above or: 
$
   \frac{1}{2}
    \left(-\text{erfc}\left(-\frac{\sqrt{9-\text{y0}^2}-2}{\sqrt{2}}\right)+\text{erfc}\left(\frac{\sqrt{9-\text{y0}^2}+2}{\sqrt{2}}\right)+2\right)
$

*So, for any given value of y0, the above is the chance x will be
big enough to make the total length greater than 3. Since we know
the probability of y=y0 (as above), the probability of the combined
events is the product of the two probabilities or:
$
   \frac{e^{-\frac{\text{y0}^2}{2}}
    \left(-\text{erfc}\left(-\frac{\sqrt{9-\text{y0}^2}-2}{\sqrt{2}}\right)+\text{erfc}\left(\frac{\sqrt{9-\text{y0}^2}+2}{\sqrt{2}}\right)+2\right)}{2
    \sqrt{2 \pi }}
$


*

*To find the total probability over all y0, we integrate the above
from -3 to +3 (since we made a special case for y < -3 and y > 3
above) numerically to get 0.211662.

*We now add this probability to the probability of |y| > 3 we
computed earlier to get 0.214362
Here's a plot of your probability function and a circle of radius 3
(note that the circle looks elongated because it follows the surface
of your probability distribution)

Note: I wrote
https://github.com/barrycarter/bcapps/blob/master/MATHEMATICA/bc-solve-stats-191040.m
to help solve this.
Former answer/hint:
Hint that's too long for a comment. Consider 3 cases: 


*

*if x < -3, the string length is > 3 

*if x > 3, the string length is > 3 

*For all other values of x, let's take x = 1.5 as an example. 

*If x = 1.5, then y^2 > 3^2-1.5^2 = 6.75, or |y| > Sqrt[6.75]. 

*Thus compute the PDF of x = 1.5 and multiple by the probability that |y| > Sqrt[6.75] (which would involve the CDF). 

*Finally, integrate.
Let me know if this doesn't help 
