Differences between wavefunction, probability and probability density? I am trying to understand the differences between wavefunction, probability and probability density. There are different definitions on the internet. 
For example:

http://inside.mines.edu/~fsarazin/phgn310/PDFs/QuantumLect4.pdf 
and

http://depts.washington.edu/chemcrs/bulkdisk/chem455A_aut05/hw_ans_nopass_wk%2004%20KEY.pdf
My questions are:
Which one is true?
What is the difference between three of them(wavefunction, probability and probability density)? 
 A: To understand the difference between probability and probability density consider the difference between mass and density.
Density is the mass per unit volume, so to find the mass you multiply the density by the volume:
$$ mass = density \times volume $$
In some cases the density will be a function of position and we have to write it as a function of the position $(x,y,z)$ so we write it as $\rho(x,y,z)$. In that case we consider a tiny cube with sides $dx$, $dy$ and $dz$, so its volume is $dV = dx \space dy \space dz$, and the mass of this tiny cube, $dm$, is:
$$ dm = \rho(x,y,z) \space dV = \rho(x,y,z) \space dx \space dy \space dz $$
And to get the total mass for some volume $V$ we have to do an integration:
$$ m = \int_V \rho(x,y,z) \space dx \space dy \space dz $$
Probability density and probability work the same way. Probability density, $P(x,y,z)$ is the probability per unit volume so the probability, ${\bf P}$ of finding our particle in the tiny cube with sides $dx$, $dy$ and $dz$ is:
$$ d{\bf P} = P(x,y,z) \space dx \space dy \space dz $$
and the probability of finding our particle in the finite volume $V$ is also calculated by integrating:
$$ {\bf P} = \int_V P(x,y,z) \space dx \space dy \space dz $$
If you look at the examples in your question they are all saying this is slightly different ways, so they are all the same. It doesn't help matters that we tend to just assume everyone knows the difference between probability and probability density and we use the same symbol, $P$, for both. In my explanation above I've used $P$ for probability density and ${\bf P}$ for probability, but we usually don't bother to make the distinction.
The probability density is calculated from the wavefunction. The wavefunction is a function of position (and time, but we'll gloss over that) so we have have to write it as $\psi(x,y,z)$, and the probability density is:
$$ P(x,y,z) = \psi(x,y,z) \psi^*(x,y,z) $$
So the probability of finding our particle in the volume $V$ is:
$$ {\bf P} = \int_V P(x,y,z) \space dx \space dy \space dz = \int_V \psi(x,y,z) \psi^*(x,y,z) \space dx \space dy \space dz $$
A: Probability is the chances of occurrence of an event, say for example the event is to find an electron around the nucleus of an atom. If the atom has an electron around the nucleus then the probability of finding the electron around the nucleus is one(1). but if the electron is to be sort some place away from the vicinity of the nucleus then the probability of finding the electron in some place away from the nucleus is zero(0).
Probability density is the term that gives the probability of occurrence of an event to be real and positive. it is understood as the probability of occurrence of the event in a certain element of given volume.the volume may be a region or an infinitesimal volume.
A: All the definitions you've posted are correct, and they aren't in conflict with each other, although they are a bit imprecise.  I'll try to explain in more detail what these concepts are.
Probability
I'll assume we don't need to attempt to define what probability is. :)  I'll just note that it's formalized in mathematics under the aegis of measure theory.
Probability Density
Probability density is a concept that naturally arises whenever you talk about probability in connection with a continuous variable, such as position of a particle.  In contrast to discrete probability (such as idealized coin-flipping or dice-rolling), we can't directly assign a probability to each individual outcome.  Probabilities have to sum to one, but there are infinitely many possible positions for a particle, so any individual position would have to have probability zero.  So how can we talk about probability spread over a continuous domain?
Well, we can talk about the probability of finding the particle within some region.  If the region covers a nonzero length/area/volume/whatever, we can expect it to have a nonzero probability.  But the exact probability will generally depend on the size/shape/position of the region in a complicated way.
Suppose we consider a very small region.  To be concrete let's suppose we're talking about a 1D position and our small region is some interval $[x, x+dx]$.  The interval has width $dx$.  It stands to reason that a small interval will have a small probability.  Moreover, if probability is "spread smoothly" in a sense that I won't bother making precise (hey, this is Physics.SE, not Math.SE!), then for small intervals the probability will be proportional to the width of the interval.  If you halve the size of the interval you get half as much probability.  Therefore
$$dP = p(x) \, dx$$
where $dP$ is the small probability of finding the particle in this small region, and $p(x)$ is some proportionality constant.  I've written it as a function of $x$ because the proportionality could vary from point to point.
Well, $p(x)$ is called the probability density, and it has the units of probability per unit length.  (Or unit area, or volume, depending on how many dimensions we do this in.)  It's a lot like mass density, which is mass per unit length/area/volume, or charge density is charge per unit length/area/volume, etc...
So, now you can imagine calculating the probability to find the particle in any arbitrary region, by slicing up that region into many small ones and summing up $dP$, which is $p(x) \, dx$, for all of them.  Well, this is just integrating:
$$P(\text{particle is in region $R$}) = \int_R p(x) \, dx$$
In particular, the particle has to be somewhere, so $p(x)$ must integrate to one:
$$P(\text{particle is somewhere}) = 1 = \int_{-\infty}^\infty p(x) \, dx$$
But $p(x)$ itself isn't limited to 1.  It isn't a probability but a probability per unit length/area/volume, so $p(x) > 1$ just means a region where probability is concentrated.  (Much like you can drive your car at 100 km/hr without actually driving 100 km.)
The Wavefunction
All of the above was just standard probability theory, nothing to do with quantum mechanics.  Now, in QM we try to predict the probability density for a particle's position (or momentum, or energy, or whatever).  We could try to do this by writing an equation for how $p(x)$ changes over time, but it turns out that doesn't give us enough information; there are situations where particles start with identical $p(x)$ but do different things as time goes on.
It's found that we do get enough information to make predictions if we write an equation for a complex-valued function $\psi(x)$, and derive the probability density from it as
$$p(x) = \psi^*(x) \psi(x)$$
The way the complex phase of $\psi(x)$ varies from point to point encodes additional information about the particle's momentum, which is necessary to predict its future behavior.  It has units of the square root of a probability density, which is a bit weird but perfectly mathematically acceptable.  This is of course the wavefunction, and the equation that determines how it varies is the Schrödinger equation.
A: It has been many years since I have used this information (so please forgive my inaccuracies, but I wanted to get you PART of the way there with an answer (since no one has responded yet).  The probability density is the integral (area under the curve) of the probability. I think the reason for the funky discrepancy between the two definitions is because..
Let's forget about wave functions for a second, and just use $x$.
If we were to integrate $x$, we would get $\dfrac{x^2}{2}$.. which can be rewritten as 
$$(.5) * x * x$$
the $.5$ can be seen as a simple scaling factor, and therefore the integral is PROPORTIONAL to 
$$x * x$$
Now, if we substitute a wave function (or probability function) in for $x$, we should get a similar result.
