Nodal Domains II by Eric J. Heller.

Let us first address a different question. The probability of finding a given particle in the union A∪B of two disjoint regions A and B equals the sum of two probabilities: the probability of finding the particle in A and the probability of finding it in B:

p(A) + p(B) = p(A∪B).

The reason this is so would be self-evident if the particle's position were sharp (that is, if it had a precise value at all times), for then the particle would be either in A or in B whenever it is in A∪B, and the sum rule p(A) + p(B) = p(A∪B) would follow from standard probability theory.

But if an electron can go through two slits without going through a single slit and without being divided in the process, then it can also be in A∪B without it being true that it is in A and without it being true that it is in B. So whence the sum rule p(A) + p(B) = p(A∪B)? To appreciate the problem, imagine two perfect detectors D_A and D_B each monitoring one of those regions. If 0 < p(A) < 1 and 0 < p(B) < 1 and p(A∪B) = 1 then

and so

Yet since p(A∪B) = 1, it is certain that either D_A or D_B will click.

Why?

You'll be surprised at the simplicity of the answer: implicit in every quantum-mechanical probability assignment is the assumption that a measurement is successfully made: there is an outcome.

So there is no mystery here. But the conclusion to be drawn is that quantum mechanics gives us probabilities with which this or that outcome is obtained in a successful measurement, not probabilities with which this or that property or value is possessed, regardless of measurements.

Thus probability 1 is not sufficient for "is" or "has".