For a given CDF compute $\Bbb P(X^2 \in A)$

Question

Let $$ F(x)=\begin{cases} 0 & \text{if} & x <-1\\ \frac{1}{4} & \text{if} & x \in [-1,0)\\ \frac{x+1}{4} & \text{if} & x \in [0,3)\\ 1 & \text{if} & x >3 \end{cases} $$

I need to get $\Bbb P(X^2 \in (\frac{1}{9},1]) $ and I got \begin{align} \Bbb P(X^2 \in (\frac{1}{9},1]) &= \Bbb P(X\in [-1,-\frac{1}{3})\cup(\frac{1}{3},1]) \\ &= F(1)-F(\frac{1}{3})+F(-1)-F(-\frac{1}{3})=\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{4} \\ &=\frac{1}{5} \end{align} Am I right?

Answer 1

It should be $F(1)-F(\frac{1}{3})+F(-\frac{1}{3})-F(-1)$, not $F(1)-F(\frac{1}{3})+F(-1)-F(-\frac{1}{3})$,

though in this particular case they happen to be te same.

Finally, $\frac{1}{2}-\frac{1}{3}=\frac{1}{6}$, not $\frac{1}{5}$.