<s>[INST] Conditional Probability Denominator doubt 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.
 [/INST] 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 $P(X=2)=\frac{1}{4}$
</s>