What's the probability a random number is at least twice as big as another? Two numbers $m,n$ are chosen from a normal distribution, i.e. the chance that either number lies between $a$ and $b$ is
$$\frac{1}{\sqrt{2\pi}}\int_a^be^{-x^2}dx$$
Edit: you could also say $e^{-x^2}/\sqrt{2\pi}$ is the probability density function.
What's the probability that $n\ge 2m$?
 A: Let $X_{1}$ and $X_{2}$ be iid random variables from $N(\mu,\sigma^{2})$ 
So you want to find 
$$P(X_{1}\geq 2X_{2})=P(X_{1}-2X_{2}\geq0)$$
where $-2X_{2}\sim N(-2\mu,4\sigma^{2})$ thus $X_{1}-2X_{2}\sim N(-\mu,5\sigma^{2})$ so converting to a standard normal with $Z=\dfrac{(X_{1}-2X_{2})+\mu}{\sigma\sqrt{5}}\sim N(0,1)$ we have 
$$=P\left(Z\geq \frac{\mu}{\sigma\sqrt{5}}\right)$$
And you can find that with standard normal table
Also note the reason why I use $X_{1}$ and $X_{2}$ is because when you say "two numbers are chosen from a normal distribution" this just means your sampling from distribution where $X_{1}$ represents first sample and $X_{2}$ represents second sample. 
A: Unless I am missing something, the answer is ${1 \over 2}$.
Note that the pdf. for each variable should either be ${1 \over \sqrt{2 \pi}} e^{- { x^2 \over 2}}$ or ${1 \over \sqrt{\pi}} e^{- { x^2 }}$ in order to
normalize to one when integrated. I will use the latter, but the answer is the same regardless (irregardless :-).
You want to compute $p= {1 \over \pi} \int_A  e^{-\|x\|^2} dx$,
where $A=\{x | x_2 \ge 2x_1 \}$. Note that $A$ is a half-space, so we can find a rotation so that $Q A = \{ x | x_2 \ge 0 \}$. The change of variables theorem
gives
$\int_A  e^{-\|x\|^2} dx = \int_{QA}  e^{-\|Q^{-1}x\|^2} dx = \int_{\{ x | x_2 \ge 0 \}}  e^{-\|x\|^2} dx $, from which we get $p= {1 \over 2}$. 
A: As mjqxxxx says, the joint distribution is circularly symmetric.  So, if you are looking for the probability that $|n|>2|m|$, then it equals the part of the circle for which $|\tan\theta|>2$  This is $4(\pi/2-\arctan2$, so the probability would be $1-(2/\pi)\arctan 2$.
