Let $X$ be a standard normal random variable. Then, $ P(X<0\mid |[X]| = 1)$ is equal to?

Question

Let $X$ be a standard normal random variable. Then, $ P(X<0\mid |[X]| = 1)$ is equal to-

1. $\frac{\Phi(1)-\frac{1}{2}}{\Phi(2)-\frac{1}{2}}$
2. $\frac{\Phi(1)+\frac{1}{2}}{\Phi(2)+\frac{1}{2}}$
3. $\frac{\Phi(1)-\frac{1}{2}}{\Phi(2)+\frac{1}{2}}$
4. $\frac{\Phi(1)-1}{\Phi(2)+1}$

My work:

First I didn't understood what is mean by $|[X]|=1$. Is it absolute value of floor of $X$? Then $$P(X<0\mid |[X]| = 1)=\frac{P(X<0, |[X]| = 1)}{P( |[X]| = 1)}.$$ Now I am stuck. Don't know how to proceed further. How can I do this? Any help would be great. Thanks.

Answer 1

Assuming $|[X]|$ means absolute value of floor of $X$, that is $1$ if and only if $X \in [-1,0) \cup [1,2)$. So $$P(X<0||[X]|=1) = \dfrac{P(X \in [-1,0])}{P(X \in [-1,0)) + P(X \in [1,2))}$$ Also useful will be symmetry, so $P(X \in [-1,0)) = P(X \in [0,1))$.