Let $X$ be a random variable with CDF: $$ F_{X}(x) = \left\{ \begin{array}{ll} 0, & \textrm{$x<-1$}\\ \frac{1}{2}, & \textrm{$-1\leq x<3$}\\ 1-e^{-x}, & \textrm{$ 3\leq x $} \end{array} \right. $$ How to compute
- $\mathbb{P}(|X-2|>1/2)$
- $\mathbb{P}(|X-2|>1)$
- $\mathbb{P}(\min(3,X)>4)$
- $\mathbb{P}(\min(3,X)<4)$
- $\mathbb{P}(\max(3,X)<4)$
EDIT: For the first one I got $$ \mathbb{P}(|X-2|>1/2) = \mathbb{P}(X-2>1/2)+\mathbb{P}(X-2<-1/2) =\mathbb{P}(X>5/2)+\mathbb{P}(X<3/2)=\\=1-1/2+1/2=1$$ Similarly second one I get $e^{-3}$. Then third probability I tried using conditional probability like this $$ \mathbb{P}(\min(3,X)>4)= \mathbb{P}(3>4|X\geq3)+\mathbb{P}(X>4|X\leq3)=0$$ Other min-max probabilities I tried the same way but I think it is wrong approach.
Remember that for all $a,b\in\mathbb R$ such that $a<b$, we have $\mathbb P(a<X\le b)=\mathbb P(X\le b)-\mathbb P(X\le a)=F_X(b)-F_X(a)$.
If the inequality is not strict on the left side, we have $\mathbb P(a\le X\le b)=F_X(b)-F_X(a-)$, where $F_X(a-)=\lim_{x\to a-}F_X(x)$.
We also have $\mathbb P(X<b)=F_X(b_-)$.
Then we have $$\begin{align*} \mathbb P(\vert X-2\vert>\frac12)&=1-\mathbb P(\vert X-2\vert\le\frac12)=1-\mathbb P(-\frac12\le X-2\le\frac12)\\ &=1-\mathbb P(\frac32\le X\le\frac52)=1-(F_X(\frac52)-F_X(\frac32-))\\ &=1-(\frac12-\frac12)=1. \end{align*}$$
$$\begin{align*}\mathbb P(\vert X-2\vert>1)&=1-\mathbb P(\vert X-2\vert\le 1)=1-\mathbb P(-1\le X-2\le1)\\ &=1-\mathbb P(1\le X\le 3)=1-(F_X(3)-F_X(1-))=1-(1-\textrm e^{-3}-\frac12)\\ &=\textrm e^{-3}+\frac12\cdot \end{align*}$$
$\mathbb P(\min(3,X)>4)=0$ since $4>3\ge\min(3,X)$.
$\mathbb P(\min(3,X)<4)=1$ since $\min(3,X)\le 3<4$.
$\mathbb P(\max(3,X)<4)=\mathbb P(X<4)=F_X(4-)=1-\textrm e^{-4}$.