When looking at the Isospin representation of a proton-neutron system (with the notation $|I,I_3\rangle$), you can go from an uncoupled to a coupled representation like this:
$$\textstyle|\frac{1}{2},+\frac{1}{2}\rangle|\frac{1}{2},-\frac{1}{2}\rangle=\sqrt{\frac{1}{2}}(|1,0\rangle+|0,0\rangle)$$
However, you can also find the following antisymmetric wavefunction when switching the first two kets around (or switching proton and neutron):
$$\textstyle|\frac{1}{2},-\frac{1}{2}\rangle|\frac{1}{2},+\frac{1}{2}\rangle=\sqrt{\frac{1}{2}}(|1,0\rangle-|0,0\rangle)$$
If you now want to calculate the probability of finding the proton-neutron system in a $|1,0\rangle$ state, you have to take the square of the inproduct of the above expressions with $\langle1,0|$, which gives $\frac{1}{2}$ for both expressions.
Do you now have to add these 2 probabilities together, since they should be equivalent, though one is symmetric and the other is asymmetric.
However, adding the probabilities means the probability to find the proton-neutron state in the $|1,0\rangle$ state is 1, but you could reason similarly for $|0,0\rangle$ and so find a total probability of 2 which doesn't seem right.
It seems weird that the wavefunction is different if you assume a proton-neutron vs. a neutron-proton system.
As a background for this question, I need to calculate the relative probabilities for 2 reactions, one of which is a pion and deuteron reacting to form a proton and neutron, so I'm writing left and right side in the coupled representation and taking the inproduct.