Find $P[2X > Y]$, normal distribution As I am learning normal distribution, i encountered this problem..

Let $X,Y$ be independent normal $N(0,1), N(1,2)$ random variables, respectively. Find $P[2X > Y]$

what got me confusing is that there is a constant in front of $X$ like for instance, if this question asked $P[X > Y]$, I could easily have $P[X-Y > 0]$ and my new normal distribution would be $N(0-1,1+2) = N(-1,3)$ but with a constant in front of $X$ i don't know what my new normal distribution is going to be Any help would be great thanks!
 A: If $X \sim \operatorname{Normal}(\mu = 0, \sigma^2 = 1)$, then $2X \sim \operatorname{Normal}(\mu = 0, \sigma^2 = 4)$, owing to the fact that $\operatorname{Var}[2X] = 2^2 \operatorname{Var}[X]$, and $\operatorname{E}[2X] = 2\operatorname{E}[X] = 2(0) = 0.$
A: Comment. It seems to me the normal density in your Comment (of an hour ago) is not right. And you answer will disagree with @Did's hint (in my experience, not a good thing). I guess you are you using the notation in which the second argument of $N(\cdot, \cdot)$ is the variance.
Consider the following simulation of a million realizations of $Z$ in
R statistical software (where the second argument of norm is the SD). Perhaps it will help you check your work. Simulated results should be accurate to a couple of decimal places.
m = 10^6;  x = rnorm(m, 0, 1);  y = rnorm(m, 1, sqrt(2))
mean(2*x > y)
## 0.341477    # aprx P(2X > Y)
z = 2*x - y
mean(z);  var(z);  mean(z > 0)
## -0.9986545  # consistent with E(Z) = -1
## 6.00205     # consistent with V(Z) = 6
## 0.341477    # aprx P(Z > 0) = P(2X > Y)
pnorm(-1/sqrt(6))
## 0.3415457   # exact answ per @Did

Below is a histogram of the simulated distribution of Z. The PDF of
$N(\mu = -1, \sigma^2 = 6)$ is superimposed.
Your answer is the area to the right of the vertical dashed line.

