Why is probability of finding the electron at a certain point when one of the slits is closed $|\Psi|^2 $ & not $|\Psi|^2 dx$? Let in a given physical condition, the wave-function to a particle be assigned as $|\Psi  (x_i,0,0,t)|^2 dx$. 
Now, at the double-slit experiment , the probability of finding the particle at any $x$ when  hole 2 is closed is given as $P_1 = |\Psi|^2$. 
My problem is here. At the first, I learnt that $P_i = |\Psi_i|^2 dx$. But in the double-slit experiment, I found only $P_1 = |\Psi_1|^2$ devoid of $dx$. Why is it so?
 A: The square of the wavefunction, $|\Psi(x,t)|^2$, is the probability density function for finding the particle. This means that the probability of finding the particle in an interval of (infinitesimal) width $\mathrm dx$ at position $x$ equals $|\Psi(x,t)|^2\mathrm dx$. On occasion, however, authors will drop the $\mathrm dx$ if it is convenient and does not affect the results they are demonstrating. This is an example of the balance one must always strike, particularly in textbook material, between clarity of exposition and thoroughness and rigour; this particular instance is probably justified as long as there is little chance of confusion (though if you want further examination of this particular usage you should state which book you are reading).
A: Given the wavefunction $\psi(x)$, the probability to find any particle within an interval $[x_0,x_0+\Delta x]$ is
$$ P([x_0,x+\Delta x_0]) = \int_{x_0}^{x_0+\Delta x} \lvert\psi(x)\rvert^2\mathrm{d}x$$
i.e. $\lvert\psi(x)\rvert^2$ is not a probability, but a probability density that has to be integrated over a set of non-zero measure to yield a notion of probability.
Formally, it is not clear what the object "$\lvert\psi(x)\rvert^2\mathrm{d}x$" is when not appearing behind an integral sign, but it may intuitively be thought of as the "probability of finding the particle in an interval of infinitesimal width $\mathrm{d}x$ around $x$".
