There is one of those thorny kinds of logic problems in the philosophy of science that goes like this:
1) The statement "All swans are white" is logically equivalent to the statement "All non-white things are non-swans", which means that if something is not white, it is not a swan.
2) Each time we find a new swan and see that it is white, that serves as evidence that indeed all swans are white.
3) By the same token, each time we find a new non-white thing and see that it is not a swan, that serves as evidence that things which are not white are not swans.  (For example, finding a red brick is evidence that things which are not white are also not swans.)
Therefore 4) each time we find a new non-white thing and see that it is not a swan, that serves as evidence that things which are swans are white, i.e., that all swans are white.
Which means that 5) finding a red brick is evidence that all swans are white.

The problem is that this logic seems to be impeccable and yet the conclusions 4 and 5 seem clearly to be false, and that is not supposed to be able to happen.  So what is wrong?

My explanation of this problem is that nothing is wrong with the logic, and that 4 and 5 are both true, but that there is a hidden psychological implication in the argument which is not true.  That implication is that each confirming instance of finding a non-white thing that is not a swan counts as much toward confirming that all swans are white as does each confirming instance of finding a white swan.  The above argument does not say that, nor does it logically imply it, which is why I call it a hidden psychological premise.

Since there are clearly many more non-white things then there are swans, sifting through all the non-white things to find a swan is not going to be as efficient in some sense as sifting through all the swans to make sure they are white.  And each confirming black crow, black cat, brown eraser, yellow pencil, ivory color computer, pink lamp, red brick, and blue piece of wrapping paper is not nearly as much evidence as finding another swan and seeing it too is white.  But that is only because the number of non-white things is so much greater than the number of swans. 

For example, suppose we need to show that all the (monthly) phone bills we received last year were less than $40. There are two ways we could do that.  We could look at each of the 12 phone bills, or we could examine all the bills we received which were $40 or more and see whether any of them were phone bills.  Suppose there are 1000 such bills.  While checking all one thousand of them and finding no phone bills would confirm that all our phone bills were less than $40, clearly each of the one thousand checks by itself only confirms that by 1/1000, whereas checking each phone bill confirms it 1/12.  And clearly it is faster and more efficient to look at the 12 phone bills than to look at the 1000 bills of $40 or more.

If we knew there were swans and if we knew that we had checked every non-white thing there was and found none of them to be a swan, we would know that all swans were white.  But since there are so many, many more non-white things than swans, each confirming instance that a non-white thing is not a swan is not as much a part of the total confirmation as is each confirming instance of finding a swan that is white.

In more general or formal terms, while it is true that:
"All A are B" is logically equivalent to "All non-B's are non-A's," it does not follow that where the number of non-B's exceeds the number of A's in a given search area that each instance of finding a non-B which is a non-A counts as much toward confirming either of the logically equivalent propositions as does finding an A which is a B.

If n_x represents the number of x's than:
1)  n_non-B > n_A 
2) each non-B that is not an A confirms both "All A's are B's" and "All non-B's are non-A's" by 1/n_non-B
3) each A that is a B confirms both "All A's are B's" and "All non-B's are non-A's" by 1/n_A
4) from 1 above, 1/n_non-B < 1/n_A
Therefore 5) each non-B that is also a non-A confirms "All A's are B's," but less than each A that is a B does.