Let $X$ be a random variable with mean $ \mu $ and variance $ \sigma ^ 2 $. Then how do we express the probability $P(|X| \ge t)$ in terms of the Q function (error function) where $t\geq 0$ is a constant.
I know that we can get this as a folded normal variable and show that $$P(|X| \ge t) = Q \left({t-\mu \over \sigma}\right) + Q \left({t+\mu \over \sigma}\right)$$
Could someone explain how to obtain this result?