Let $X$ be a random variable with the following cumulative distribution function:
$$F(x)= \begin{cases} 0 & \quad x<0\\ x^2 & \quad 0\leq x<\frac{1}{2}\\ \frac{3}{4} & \quad \frac{1}{2}\leq x<1\\ 1 &\quad x\geq1. \end{cases}$$
Then what is the value of $P(\frac{1}{4}<X<1).$
I am trying to solve this problem as:
$P(\frac{1}{4}<X<1)=P(\frac{1}{4}<X\leq1)$, since for continuous random variable probability at a point is always zero. Thus $P(\frac{1}{4}<X<1)=P(\frac{1}{4}<X\leq1)=F(1)-F(\frac{1}{4})=\frac{15}{16}=0.9375$. But the answer didn't match. Answer is 0.68. So where did i wrong. Any suggestion or solution regarding this should be highly appreciated.
$$P\left(\dfrac{1}{4}<X<1\right)=F\left(1^-\right)-F\left(\dfrac{1}{4}^{+}\right)=\dfrac{3}{4}-\dfrac{1}{16}=\dfrac{11}{16}$$