Conditional Probability Denominator doubt

Question

Let X be a rv such that $P(X = 2) = 1/4$ and its CDF is given by $$F_{X}(x) = \begin{cases} 0,& x< -3\\ \frac{3}{4}(x+3),& -3\leq x <2 \\ 3/4,& 2 \leq x <4\\ \frac{3}{64} x^2,& 4 \leq x < \frac{8}{\sqrt{3}} \\ 1,& x \geq \frac{8}{\sqrt{3}} \end{cases} $$

$x=2$ in the only discontinuity of $F$. I have to compute $P(X<3|X \geq 2)$.

My way is: $\frac{P(2 \leq X<3)}{P(X \geq2)}$ = $\frac{F(3)- F(2) - P(X = 3) + P(X=2)}{1 - P(X<2)}$

$P(X= 3) = 0$ because it is continuous there. How can I compute the denominator.

Answer 1

$P(X>2)=\int_4^{8/\sqrt3 } \frac 3 {32} xdx=\frac 1 4$. Add $P(X=2)=\frac 1 4$ to this to get $P(X \geq 2)$.

Answer 2

Having the CDF no integral is needed.

You can solve the problem in this way

$$\mathbb{P}[Z<3|X\geq 2]=\frac{F_X(3)-F_X(2^-)}{1-F_X(2^-)}=\frac{\frac{3}{4}-[\frac{x+3}{10}]_{x=2}}{1-[\frac{x+3}{10}]_{x=2}}=\frac{1}{2}$$

The current CDF is wrong. It cannot be $\frac{3}{4}(x+3)$ when $-3 \leq x <2$ because it is $F(2^-)>1$

The previous was right...$F_X(x)=\frac{x+3}{10}$, according with $P(X=2)=\frac{1}{4}$