There are no negative probabilities.  

There exists in the [phase space formulation][1] of QM “quasi-probability” distributions $W(x,p)$ that are *somewhere* negative, but the probabilities that occur as a result of integrating such distributions are always non-negative.

To be explicit, 
$$
\vert\psi(x)\vert^2=\int dp\, W(x,p)
$$
(for instance) is obtained by integrating all momenta of the joint quasi-distribution $W(x,p)$, $\vert \psi(x)\vert^2$ agrees with the probability density computed from the Born rule, and is everywhere non-negative so that the probability of finding the system in some space interval is *always* non-negative.

To illustrate this, look at the following two figures: 

[![enter image description here][2]][2]
[![enter image description here][3]][3]

They show the same quasi-distribution function, which contains regions of negativity near the center, as illustrated in the right figure: the near the central "anti"-peak the quasidistribution is clearly negative.  However, integrating this quasi distribution along the line $p$ at $x=-2$ (shown on the left figure by the red line) yields

$$
\int dp W(x=-2,p)=\vert\psi(x=-2)\vert^2 > 0 
$$
so even if there are regions of negativity the probability of finding
the system near $x=-2$, which is $\vert\psi(x=-2)\vert^2\,dx$, is of course positive.  For completeness here $\psi(x)$ is in fact $\psi_1(x)$, the wavefunction of the $n=1$ harmonic oscillator state.

To make absolutely explicit the difference between a probability *density* and a probability, consider the following normalized Gaussian wavefunction:

$$
\psi(\sigma)=\frac{e^{-x^2/(2\sigma)}}{(\pi \sigma)^{1/4}} \tag{1}
$$
for $\sigma=1/8$.   $\psi(1/8)$ is a perfectly legitimate wavefunction for a particle in a harmonic oscillator.  The resulting probability *density* $\vert \psi(1/8)\vert^2$ is $>1$ near the origin:

[![enter image description here][4]][4]

but of course this is not a problem.  Since $\int_{-\infty}^{\infty} dx \vert \psi(1/8)\vert^2=1$ and $\vert \psi(1/8)\vert^2 \ge 0$ everywhere, it follows that, for any subinterval $[a,b]$: 
$$
\int_a^b dx \vert \psi(1/8)\vert^2 \le 1 
$$
even if $\vert \psi(1/8)\vert^2$ can be greater than one *somewhere*.  Nobody in their right mind would suggest that, because the probability *density* $\vert \psi(1/8)\vert^2$ is greater than 1 *somewhere*, probabilities greater than 1 are possible possible.  

It's the same argument for negative regions of Wigner functions, which are (quasi)probability *densities*, with the distinction that these (quasi)probability densities can be negative.


  [1]: https://en.m.wikipedia.org/wiki/Phase-space_formulation
  [2]: https://i.sstatic.net/olhA9m.png
  [3]: https://i.sstatic.net/xS86fm.png
  [4]: https://i.sstatic.net/q0xUam.png