Probability mass function and Probability density function What is the Difference between probability Mass function and probability density function?
Why the value of continuous probability distribution function is not the probability for particular input point for example for a continuous distribution say(assume it is pdf under suitable domain).
f(x) = x^3 why it is not true that f(3) is probability at 3.
what do you mean by f(3)?
 A: Informally, you can think of it as $f(3)$ simply returning the height of the density function, but this is really of no interest. Recall that $f(3)=Pr(X=3)$, and in our case $X$ is a continuous random variable. It does not make sense to speak of probabilities at certain values of $X$ because of this. 
For example, let's say I toss a ball. It lands around a 20 feet mark. But, in fact, when I look closer it was more like 19.75. But, looking even closer, it was more like 19.745. But, looking even closer than that, it was more like 19.7445. And so we could continue... It doesn't end! Hence, there is no probability for a certain distance of my toss and we use intervals instead. That is, the probability that I'd toss the ball 20 feet is 0 ($f(20)=Pr(X=20)=0$), but the probability that it's between some interval is not ($Pr(20-\delta < X < 20 +\delta)\neq 0$).
Maybe this link can be helpful. 
A: I stumbled upon this question and found my answer in MIT OCW notes (Reading 5b and 4a to be precise). I'm quoting the explanation from there.

Probability mass and probability density - these terms are completely analogous
to the mass and density you saw in physics and calculus.
Mass as a sum: If masses m1, m2, m3, and m4 are set in a row at positions x1, x2, x3, and x4, then the total mass is m1 + m2 + m3 +
m4.
We can define a ‘mass function’ p(x) with p(xj ) = mj for j = 1, 2, 3,
4, and p(x) = 0 otherwise. In this notation the total mass is p(x1) +
p(x2) + p(x3) + p(x4). The probability mass function behaves in
exactly the same way, except it has the dimension of probability
instead of mass.
Mass as an integral of density: Suppose you have a
rod of length L meters with varying density f(x) kg/m. (Note the units
are mass/length.)
If the density varies continuously, we must find the total mass of the
rod by integration:
total mass $ = \int_{0}^{L} f(x)  dx$
This formula comes from dividing the rod
into small pieces and ’summing’ up the mass of each piece. That is:
total mass ≈ $\sum_{i=1}^{n} f(x_i) = \Delta x$ In the limit as $\Delta x$ goes to zero the sum
becomes the integral. The probability density function behaves exactly
the same way, except it has units of probability/(unit x) instead of
kg/m. Indeed, equation (1) is exactly analogous to the above integral
for total mass. While we’re on a physics kick, note that for both
discrete and continuous random variables, the expected value is simply
the center of mass or balance point.

Reference (I couldn't figure out to use "Insert citation" but doing so is required from MIT OCW terms, hence I am manually inserting it here):
Jeremy Orloff and
Jonathan Bloom. 18.05 Introduction to Probability and Statistics.
Spring 2014. Massachusetts Institute of Technology: MIT OpenCourseWare, https://ocw.mit.edu/. License: Creative Commons BY-NC-SA.(Terms : https://ocw.mit.edu/terms/)
