Let $X$ be a random variable with the following cumulative distribution function $$F(x) = \begin{cases} \ {0}, & x<0 \\[2ex] \ x^2, & \ 0\le x<\dfrac{1}{2}\\[2ex] \ \dfrac{3}{4}, &\dfrac{1}{2}\le x< 1 \\[2ex] \ 1 & x\ge1 \end{cases}$$
Find $P(\dfrac{1}{4}<X<1)$
My input
Here I used $P(X\le x)= F(x) $ and $P(X< x)= F(x)^{*}$
At $\dfrac{1}{2}$ our pdf is discontinuous and in the interval $0\le x<\dfrac{1}{2}$ $f(x)$ is continuous and in $\dfrac{1}{2}\le x< 1 $ it is discrete. So
$P(\dfrac{1}{4}<X<\dfrac{1}{2})+P( \dfrac{1}{2}\le x< 1 )=F\bigg(\dfrac{1}{2}\bigg)^{*}-F\bigg(\dfrac{1}{4}\bigg)+F(1)^{*}-F\bigg(\dfrac{1}{2}\bigg) + P\bigg(X=\dfrac{1}{2}\bigg)$
$= F\bigg(\dfrac{1}{2}\bigg)^{*}-F\bigg(\dfrac{1}{4}\bigg)+F(1)^{*}-F\bigg(\dfrac{1}{2}\bigg) + F\bigg(\dfrac{1}{2}\bigg)-F\bigg(\dfrac{1}{2}\bigg)^{*}$
$=F(1)^{*}-F\bigg(\dfrac{1}{4}\bigg)=\dfrac{11}{16}$
So this question came in one mark. Is there any quick way to solve this ?
$$P\left(\frac14<X<1\right)=P(X<1)-P\left(X\leq\frac14\right)=\lim_{x\to1-}F(x)-F\left(\frac14\right)=\frac34-\frac1{16}=\frac{11}{16}$$