How do you come up with a POVM? This is a made-up example, just to understand a concept. If changing the probability values aids your explanation, that's fine by me.
Say you have a physical quantity $E$ that can take values 1, 2, 3 with probabilities 0.4, 0.25, 0.35 respectively (working in a quantum framework). You have a positive operator valued measure, $E_1, E_2, E_3$, with $E_i$ corresponding to your measurement resulting in value $i$. If $\rho$ is the density operator representing the current state, then you have:
Tr($\rho E_1$) = 0.4, Tr($\rho E_2$) = 0.25, Tr($\rho E_3$) = 0.35
Given just these probability values, is it possible to construct $E_1, E_2, E_3$ "backwards"?
 A: Yes. Not only is it possible, the solution is not unique and there exists a solution that is independent of $\rho$. 
So for $p_k$ being probabilities with $\sum_k p_k = 1$, you can say $E_k = p_k \mathbf{1}$, where $\mathbf{1} = \sum_j |j\rangle \langle j|$ (ie the identity matrix). Then $Tr[ E_k \rho] = p_k Tr[\rho] = p_k$. This works for all $\rho$.
There are also nontrivial solutions. If say your state is a pure state $|\psi\rangle$ in some $d$ dimensional Hilbert space then you only need to find the basis $|k\rangle$ for which $|\psi\rangle = \sum_k \sqrt{p_k} e^{i \phi_k} |k\rangle$. 
Also notice something with the probability equations, in $Tr[\rho E_1]$ if you don't assume that $\rho$ is the state it could well be $E_1$ is the state and $\rho$ is the measurement (although unlike $\rho$, $E_k$ need not be normalized). So the question is the same as asking how to construct three states for which measurement of $\rho$ gives the said probabilities. 
A: If you count the number of variables that you want to find you get 18 (=9+9, 2 hermitian operators, the third is fixed by the others), but you impose only four conditions on these variables (=2x2, the trace may be complex, the third trace is already counted), therefore I think it's impossible.
