Whenever I try...

Well...not quite.
If you are talking about 'passing' in the sense that you get a grade of 65 and up, then yes, you're absolutely correct. Naturally, this does not apply if the class has a Belle Curve Adjustment applied to it such as in some college classes...such as the class where you learn how to make a Belle Curve. These classes can have a 50 as a C if that's the average.
But I digress.
If you are talking about whether or not a given answer will be correct is based on whether or not the previous answer(s) are correct is not correct. This can be shown using a simple coin flip. The amount of heads should, roughly, equal the amount of tails. Granted, there is usually some deviation is usually in place, but as the amount of coin flips increases, it should get closer and closer to 50%.

The same applies to T/F questions. They will always lean towards 50%, and subsequently not depend on whether or not a previous question is correct.
Furthermore, it does not "randomly" decline. It would, instead, approach a certain number depending on the number of answers to a given question. A multiple choice true or false question would approach 25% for example. T/F would approach 50%.
A mixture of the two would result in
(50%+25%)/2= 37.5% assuming the test is half T/F and half four choice multiple choice.
If there is an uneven amount of questions, then we would need to model things differently.
It can actually be modeled using sigma notation for further simplification, if you'd like me to write it out.

tl;dr
You're wrong.

but...
100/4 = 25 right?
100/8 =12.5 right?
now, with 2 questions, there would be two answers that are correct,
12.5 * 2 = 25...

Im just saying... in my head this sounds right... But go ahead and prove me wrong :)

and, i might be thinking in a completely retarded way here, but idgf...

Let's pretend there is a two question multiple choice test...
each question has 4 possible answers, and the test taker is guessing (= randomness)

Question one has a 25% chance of being right with a 75% chance of being wrong
question two has the same principles, but now that there are two questions with eight possible answers and only two correct answers, the problem does not double... instead it multiplies!

Consider that out of sheer luck the first answer is correct. Well That's only 50% of the test! Now the remaining 50% still has to face a 1 in 4 chance to acheive a passing grade!

If one were to guess an entire test... I would be willing to bet that a 4 part multiple choice would eventually even out with a T/F exam if only enough questions were either asked or enough answers were given... Of course randomness would be the common denominator and not a single factbased or educated guess would sustain an "answer".

en.wikipedia.org/wiki/Permutations

If you learn about permutations you will do a lot of word problems to do with testing and answer key arrangements.

Okay. Thank you for enlightening me :)