X and Y are two independent standard gaussian variables, what is the Probability that X>100*Y? We want to find $P(X> 100*Y)$.
My strategy was to first express $X>100*Y$ as $X-100*Y>0$ and express $Z=X-100*Y$.
$Z$ should be another normal random variable, yet not standard. My approach further was to standardise it so I can use 68–95–99.7 rule. However this approach is leading nowhere.
Can anyone give any hints how I should approach it?
 A: I really like your method of using $Z = X - 100\cdot Y$
You were definitely correct in pointing out $Z$ is again Normal! We can characterize every normal distribution by its mean and variance. So to find out $Z$ exactly we just need these two things.
Mean of Z
You will have seen before that expectation is "linear" that is we know $\mathbb{E}[Z] = \mathbb{E}[X] - 100\cdot \mathbb{E}[Y] = 0 - 100\cdot 0 = 0$
Variance of Z
This is slightly harder, in general variance doesn't add up nicely but because $X$ and $Y$ are independent then it will! 
Var$[Z] = $ Var$[X] + 100^2 $Var$[Y] = 1+10,000 = 10,001$
Finishing off
Hence $Z \sim N(0,10001)$ 
And because Normal distributions are symmetric about their mean we know $\mathbb{P}[Z > 0 ] = \frac{1}{2}$
(Yes we didn't need to calculate the variance at all)
Another approach
I first saw this question a while back and used a different method. Yours I think is much better but I will include mine here.
We find the probability of $X>100\cdot Y $ by conditioning on $4$ events, namely the signs of the two random variables. Notice that as $X$ and $Y$ are standard normal they both have equal probability of being positive or negative.

*

*$X>0$ , $Y<0$. In this case we know for certain $X>100Y$

*$X<0$, $Y > 0$. In this case we know for certain $X < 100Y$

*$X>0$, $Y>0$. Let $p$ be the probability of $X>100Y$ when they are both positive.

*$X<0$ , $Y<0$. By symmetry the probability of $X>100Y$ is $1-p$
Hence $\mathbb{P}[X>100Y] = \frac{1}{4}(1+0+p+1-p) = \frac{1}{2}$
A: Suppose $X $ and $Y$ are independent random variables with $$ X\sim N(\ \mu_x,\ \sigma^2_x\ ),\quad Y\sim N(\ \mu_y,\ \sigma^2_y\ ).\ $$
Then,
$$-100Y \sim N(\ -100\mu_y,\ 100^2 \sigma^2_y\ )\ ,\quad $$
and so
$$X-100Y = X + (-100Y) \sim N(\ \mu_x - 100\mu_y,\ \sigma^2_x\ + 100^2 \sigma^2_y\ ). $$
Therefore,
$$P(X>100Y) = P(X-100Y>0) = \ldots$$
