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By convention$ P[X=x] = 0$ for all x. How would you explain probability density function $f(x) = 3x^2$ (where x is between 0 and 1), probability is 0 otherwise. Then when x =0.9. f(x) > 1 which does not equal to 0

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    @user133466: Chris has given the same answer$I$would have given. If $h$ is small, then the probability that $0.9\le x\le 9+h$ is about $3(0.9)^2h$.2012-10-11

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When you have a probability density $f_X$ describing the distribution of a random variable $X$, you have $P(X=x) = 0$ for any $x$. There are not point-masses in such a distribution. In fact, any denumerable subset of the line has probability zero.

Such a function renders this service. $P(A) = \int_{A} f_X(x)\, dx $ for a decent(measurable) set of real numbers.

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    @DilipSarwate that's certainly fair and a useful approximation for continuous densities. Trouble starts popping up if you use stuff like that with discontinuous densities (for example piecewise constant densities) or if you (like me) tend to work with things that are only defined a.e. Your response is probably better for him though :)2012-10-11