Quantum mechanics electron probability
The issue here is that Schrödinger's equation doesn't yield us a &quot;solution of 90% probability&quot;. It doesn't yield a probability at all. It yields a wave function, $\psi(x,y,z,t)$, which has units $\frac{1}{\sqrt{m^3s}}$. To get a probability density, we square its norm: $|\psi|^2=\psi^*\psi$. Note: even this is not a probability. A probability is dimensionless, whereas this has units $\frac{1}{{m^3s}}$. We can represent the probability of the particle described by $\psi$ being in the range (x, y, z, t) and (x+dx, y+dy, z+dz, t+dt) by multiplying the norm squared by dxdydz*dt: $P(x,y,z,t) = |\psi(x,y,z,t)|^2dxdydzdt$. Of course, these increments are infinitesimal, so this alone would evaluate to zero. However, this approach allows you to integrate over a range of x,y,z,and t:
$P(x+\Delta{x},y+\Delta{y},z+\Delta{z},t+\Delta{t}) = \int{dx\int{dy\int{dz\int{dt|\psi(x,y,z,t)|^2}}}}$.
A better understanding of this will help you understand the issues with your second question. However, we can also rely on the Pauli exclusion principle to assert that there cannot be two Fermions that share the same set of quantum numbers (note that some of the quantum numbers may be shared, but not all simultaneously.