# What is the probability of a continuous uniform random variable in $[0,1]$ to be $1/2$?

Suppose I have a random variable $x$ with uniform pdf over $[0,1]$. When I evaluated the probability $P(x=1/2)$, I got 0. How can that be correct? After all the probability exists and is equally likely for any value from 0 to 1 , right?

$P(x=1/2)= \lim_{\epsilon \rightarrow 0} \int _{(1/2 )-\epsilon} ^{(1/2 )+\epsilon} f(x)dx = 0$

where $f(x)$ is the pdf. After all what is the probability $P(x=1/2)$?

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You're correct. Any absolutely continuous random variable $X$ has $P(X=x)=0$ for all $x$. What makes the uniform distribution "uniform" is that $P(X\in [a,b])=b-a$ for all $0\leq a<b\leq 1$, i.e. the probability that $X$ falls within an interval only depends on the length of the interval. In particular $P(X\in [0,\tfrac14])=P(X\in [\tfrac14,\tfrac12])$. – Stefan Hansen Oct 2 '13 at 12:19
This does touch upon the finer aspects of infinitesimals. No line has an area, but intuitively, any integral is the sum of the areas of infinitely many lines. In exactly the same way, the probability of getting any single outcome is zero, but the probability of getting an outcome in a rangle is equal to the sum of all zeros within that range. This is counterintuitive, but that's the way calculus works. – Arthur Oct 2 '13 at 12:26

It is $0$.

You seem to think every possible event must have non-zero probability. That's not true for continuous probability distribution. For a reference consult your textbook, or see this wiki page.

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$P(x=1/2)= \lim_{\epsilon \rightarrow 0} \int _{(1/2 )-\epsilon} ^{(1/2 )+\epsilon} f(x)dx = \lim_{\epsilon \rightarrow 0} \int _{(1/2 )-\epsilon} ^{(1/2 )+\epsilon} 1dx = 0=\lim_{\epsilon \rightarrow 0} 2 \epsilon = 0$