Actually, you can solve a fair bit of this using logic, without too much stats knowledge.
In reality, there are seven cases to consider: if you study 0, through if you study all 6. If you study 0 or 1, you're not gonna know at least 2 answers. If you study 5 or 6, well, you can always not answer one of the questions, so that's gonna be 100%. So you just need to figure out 2, 3, and 4.
Now, 3 of the 6 questions will be chosen for the test. Let's say you studied 3. Since both are randomly distributed, it stands to reason that the chance of 2 or 3 matching (total) is the same as the chance of 0 or 1 matching (total). If "it stands to reason" is not a convincing enough argument, consider that for any three questions chosen, the opposite three will have an equal chance of showing up, and thus contribute to the opposite total. Thus, for 3, it's 50%.
The tricky part is figuring out 2 and 4 based on logic. In this case you will have to use some elementary probability. Let's take 2 first. You have to guess 2 questions to study for, and both have to actually appear on the test. Well, when you guess the first one, you have a 50/50 chance of being right (3 test questions out of 6 total). In order for this to count, you have to guess it right. This means that there's 2 test questions left, out of 5 questions total. When you guess the second time then, you thus have a 40% chance of being right. Thus, 50% * 40% = 20%. So you have a 20% chance of having both questions appear on the test.
And as with before, you can just "flip" all the logic for the guessing 2, to figure out that guessing 4 means 80% chance of both appearing. After all, guessing 4 and hoping that at least 2 will appear, is equivalent to guessing which 2 won't appear, and being right on at least one of them. Since guessing which 2 won't appear and being WRONG on both of them has a 20% chance of happening (via the above), the former has an 80% chance of happening.
Of course, these percentages have to be decreased to take into account the Murphy coefficient, and increased if a bribe coefficient exists.
Chuck Hsiao
Amptron
Edit: Tried to clarify my logical argument for 4 questions studied. This whole thing with the text box scrolling instead of wrapping makes these logical arguments really hard to read.
