Probability of getting a higher value from a normal distribution You have two independent random variables $X$ and $Y$ each of a zero mean, unit variance distribution. What is the $P(X>5Y)$?
 A: Here are some hints to answer the question:


*

*$\mathbb E(Z)=\mathbb E(X-b\cdot Y)=\mathbb E(X)-b\cdot \mathbb  E(Y)$

*The linear combination of two (or more) normally distributed variables is normally distributed as well.

*In this case we don´t need to know the variance of $Z$. We can use the property of symmetry of the normal distribution. 
A: Perhaps surprisingly, the answer is $1/2$.  Aside from callculus's approach, there is a short-circuit-y way to see the same thing.  For any given values $X$ and $Y$, we can observe that $X > 5Y$ if and only if $-X < 5(-Y)$.  By symmetry*, then, the probability is $1/2$.
*We have to be careful here, because $X$ and $-X$ have to be "equally likely" (in a way that can be made precise), as well as $Y$ and $-Y$.  Fortunately, that happens to hold in this case.

Here is a pictorial representation of the situation.  The red line is $X = 5Y$, and the blue blur is a schematic depiction of the normal bivariate ordered pair $(X, Y)$.  How much of that blur is below the line, where $X > 5Y$?  How much of it is above the line, where $X < 5Y$?

Hopefully, this makes things more intuitively satisfying.  Note that the graphical approach allows you to answer such questions as what the probability is if we require $Y > 0$.  In that case, we restrict ourselves to the upper half-plane, and since the bivariate normal distribution is radially symmetric, we see that the probability is just $\frac{\arctan 1/5}{\pi} \approx 0.062833$.  (Oddly, this is very close to, but not equal to, $\pi/50$.)

Finally, some quick code:
import random

length = 1000000

random.seed()

tally = 0
for i in range(length):
    x = random.gauss(0, 1)
    y = random.gauss(0, 1)
    # while y < 0:
    #     y = random.gauss(0, 1)
    if (x > 5*y):
        tally += 1

print(float(tally)/length)

The code that is commented out forces y to be positive.  If we leave it commented out, we get
% python normal.py
0.500236

If we put the code back in, we get
% python normal.py
0.062806

Just a couple of data points.
