Probability based on cumulative distribution function I have the following cumulative distribution function:
$F(x)=\begin{cases}
 0& \text{ if } x<0 \\ 
\frac{1}{4}+\frac{1}{6}(4x-x^2) & \text{ if } 0\leq x<1\\ 
 1& \text{ if } x\geq1
\end{cases} $
Now, it is required to calculate the probability $P(X=0|0\leq x<1)$. How can I obtain the probability at a single point when the function given is a continuous function in the given range $ 0\leq x<1$?
Edit: Actually, I know how to solve these types of questions but I am getting problem in this problem specifically. I think that the condition that has been given in this probability makes the density continuous and the probability will turn out to be zero. If I had to solve the problem without condition ($ 0\leq x<1$), then probability that the random variable takes value zero will be equal to $\frac{1}{4}$. But, for this problem answer has been given as $0.33$. I am not able to understand how this answer has been obtained?.
 A: We have $$\mathbb{P}[X=0|0\leq X<1] = \dfrac{\mathbb{P}[X=0,0\leq X<1]}{\mathbb{P}[0\leq X<1]}$$ $\{X=0\}\cap\{0\leq X<1\} = \{X=0\}$ 
so the denominator above is equal to $\mathbb{P}[X=0]$, and $\mathbb{P}[0\leq X<1] = F(1)-F(0) = \frac{3}{4}$.
However, it should be noted here that this CDF has two jump discontinuities at points $x=0$ and $x=1$ as seen here:

This means that $X$ is in fact a mixed random variable, and since CDFs are right-continuous, we have that $\mathbb{P}[X=0]=\frac{1}{4}$, which is the size of the jump at $x=0$. Thus, our desired probability is equal to $\frac{1}{3}$.
On a last note, while this answer might not provide much more information about the result itself, I thought it informative to mention the fact that $X$ is a mixed r.v. (thus "both" discrete and continuous) and that the value at $x=0$ (or any other discontinuity point for that matter) should not be taken lightly, or at least, without any mention to the properties of the CDF.
A: 1/4 is the place to start.  However, you have to divide by the probability that x is less than 1 (the given).  P(X<1) = 0.75 or 3/4.  So 1/4 divided by 3/4 gives you 1/3.
