Why can we predict any experimental outcome, given a probability density over quantum states? There is a very interesting answer given by Peter Shor in this website here. However, I admit I don't fully understand it.
In particular, I don't understand:

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If we have a probability density μv on quantum states v, we can predict any experimental outcome from the density operator

What is the formula given? Is it $\operatorname{trace}{\left(\rho A\right)}$ for an observable $A$?


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a probability distribution on quantum states is an overly specified distribution, and it is quite cumbersome to work with

How can we work with the density operator if, from an experimental point of view, all we can measure is the probability distribution?
 A: From your question, it sounds like you've learned about observables but maybe not von Neumann measurements and POVMs. In my previous answer, when I mentioned "experimental outcome", I was referring to these types of measurement. If you want to understand my previous answer completely, you probably should learn about them.
For your first question, the expectation valuable of an observable $A$ on a density matrix $\rho$ is indeed $\mathrm{Tr}\, (\rho A)$. 
For the second question, if you can prepare a mixed quantum state (represented by a density matrix) repeatedly, you can measure the density matrix by doing quantum state tomography on it. If you only have one instance of a quantum state, you can't even get a probability distribution by measuring it ... you just get one measurement outcome. 
The probability distribution I referred to in my previous answer is a probability distribution that the OP was asking about. This is related to how the state was prepared, and you can't measure this probability distribution even if you can prepare arbitrarily many copies of the quantum state.
