Computing Probability given CDF

Question

Given $$F(x) =\begin{cases}(x-1)^2/4,& 1 < x < 3\\0,& x\le 1\\ 1,& x\ge3\end{cases}$$

Is $P(2 < X < 5) = P(2 < X)$ since $F(x) = 1$ when $x\ge 3$? Also would $P(X = 1.5) = f(1.5) = F'(1.5)$?

Answer 1

We have $$P(2<X<5) = F_x(5)-F_x(2) = 1 - F_x(2) $$ On the other hand, $$P(2<X) = 1-P(X<2) = 1 - F_x(X<2)$$ So the first answer is yes.
But when we are talking about one specific point in a continuous distribution, its probability considered zero as its occurrence will be almost impossible. Assume we have $$ P(X=1.5) = P(1.5-\epsilon < X < 1.5+\epsilon) = F_x(1.5+\epsilon) - F_x(1.5-\epsilon) $$ where $\epsilon$ is very small number. Based on Taylor series expansion $F_x(1.5+\epsilon)$ could be written as $$ F_x(1.5+\epsilon) = F_x(1.5) + f_x(1.5)(\epsilon) = F_x(1.5) + \epsilon $$ and $F_x(1.5-\epsilon) = F_x(1.5) - \epsilon $ in the same way. Therefore, $$ P(X=1.5) = F_x(1.5) + \epsilon - F_x(1.5) + \epsilon = \epsilon \approx 0 $$ I hope it helps.