Nay. The question has no correct answers. Nor does it lead to any kind of contradiction, though it may seem like it. Lets work at it. 

If you ask about picking something at random from a set, you need to know how many times it appears, and how many total elements are in the list. In other words, you need to know the proportion of that given something in a list. 

So, we can evaluate the odds of picking a certain answer at random by dividing the amount of times it appears over the total amount of answers, and multiplying by 100 to get a percent value. If this value is indeed what the answer states, only then is the answer correct.

Lets write down an equation to describe this a little better.

(i/a) * 100 = x ; where x is the percentage given in the answer, i is the amount of times x appears in the list, and a is the total number of elements in the list.

If you plug in the numbers, and the result is some True mathematical statement, then that is indeed the correct answer. 

So, if the question were the same, but the answers were:

A)50%
B)25%

Then the answer would safely be A, as it indeed is 50% of the list. (1/2 * 100 = 50% is True)

Now, if we look at the actual answers:

A)25%
B)50%
C)60%
D)25%

And then looked at how much of each question constitutes space in the list:

A)50% (A and D are equivalent, so we can say it appears twice and omit D.)
B)25%
C)25%

We find that none of the answers has a value equal to the proportion of which they show up in. Therefore, none of them are correct answers, and the Professor is a sneaky bastard for posing this question. Q.E.D.