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We propose a solution for the apparent paradoxof the double-slit experiment within the framework of our reconstruction of quantum mechanics (QM), based on the geometrical meaning of spinors. We argue that the double-slit experiment can be understood much better by considering it as an experiment wherebythe particles yield information about the set-up rather than an experiment whereby the set-upyields information about the behaviour of the particles. The probabilities of QM are conditional, whereby the conditions are defined by the macroscopic measuring device. Consequently, they are not uniquely defined by the local interaction probabilities in the point of the interaction. They have to be further fine-tuned in order to fit in seamlessly within the macroscopic probability distribution, by complying to its boundary conditions.When a particle interacts incoherently with the set-up the answer to the questionthrough which slit it has moved is experimentally decidable.When it interacts coherently the answer to that question is experimentally undecidable.We provide a rigorous mathematical proof of the expression $\psi_{3}= \psi_{1} + \psi_{2}$ for the wave function $\psi_{3}$ of the double-slit experiment,whereby $\psi_{1}$ and $\psi_{2}$ are the wave functions of the two related single-slit experiments. This proof is algebraically perfectly logical and exact, but geometrically flawed and meaningless for wave functions. The reason for this weird-sounding distinction is that the wave functions are representations of symmetry groups and that these groups are curved manifolds instead of vector spaces. The identity $\psi_{3}= \psi_{1} + \psi_{2}$ must therefore be replaced in the interference regionby the expression$\psi'_{1} + \psi'_{2}$, for which a geometrically correct meaning can be constructed in termsof sets (while this is not possible for $\psi_{1} + \psi_{2}$). This expression has the same numerical value as $\psi_{1} + \psi_{2}$ , such that $\psi'_{1} + \psi'_{2} = \psi_{1} + \psi_{2}$,but with $\psi'_{1} = e^{\imath \pi/4 } (\psi_{1} +\psi_{2})/\sqrt{2} \neq \psi_{1}$ and $\psi'_{2} = e^{-\imath \pi/4 } (\psi_{1} +\psi_{2})/\sqrt{2} \,\neq \psi_{2}$. Here $\psi'_{1}$ and $\psi'_{2}$ are the correct (but experimentally unknowable)contributions from the slits to the total wave function $\psi_{3} = \psi'_{1} + \psi'_{2}$. We have then $p = |\psi'_{1} + \psi'_{2}|^{2} = |\psi'_{1}|^{2} + |\psi'_{2}|^{2} = p'_{1}+p'_{2} $ such that the apparent paradox that quantum mechanics would not follow the traditional rules of probability calculus for mutually exclusive events disappears.