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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?.

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    $\begingroup$ This self-study question requires more input from you on why you cannot solve it on your own. $\endgroup$ – Xi'an Jan 19 '18 at 11:29
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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: enter image description 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.

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  • $\begingroup$ One more thing, can you suggest me any source from where I can find more of such mixed CDF type problem. $\endgroup$ – user8125394 Jan 20 '18 at 2:14
  • $\begingroup$ The link I provided above is a good and quick intro, but if you are looking for more detailed explanations and examples you can find some here and here. $\endgroup$ – Emil Jan 20 '18 at 11:33
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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.

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  • $\begingroup$ F(1) = 1 according to the formula $\endgroup$ – aginensky Jan 19 '18 at 14:07

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