Exam with $12$ yes/no questions (half yes, half no) and $8$ correct needed to pass, is it better to answer randomly or answer exactly 6 times yes?
We are given the fact that there are $12$ questions, that $6$ have the correct answer "yes" and $6$ have the correct answer "no."
There are $\binom{12}{6} = 924$ different sequences of $6$ "yes" answers and $6$ "no" answers. If we know nothing that will give us a better chance of answering any question correctly than sheer luck, the most reasonable assumption is that every possible sequence of answers is equally likely, that is, each one has $\frac{1}{924}$ chance to occur.
So guess "yes" $6$ times and "no" $6$ times. I do not care how you do that: you may guess "yes" for the first $6$, or flip a coin and answer "yes" for heads and "no" for tails until you have used up either the $6$ "yeses" or the $6$ "noes" and the rest of your answers are forced, or you can put $6$ balls labeled "yes" and $6$ labeled "no" in an urn, draw them one at a time, and answer the questions in that sequence.
No matter what you do, you end up with some sequence of "yes" $6$ times and "no" $6$ times. You get $12$ correct if and only if the sequence of correct answers is exactly the same as your sequence. That probability is $\frac{1}{924}.$
There is no way for you to get $11$ correct. You get $10$ correct if and only if the correct answers are "yes" on $5$ of your "yes" answers and "no" on your other "yes" answers. The number of ways this can happen is the number of ways to choose $5$ correct answers from your $6$ "yes" answers, times the number of ways to choose $5$ correct answers from your $6$ "no" answers: $\binom 65 \times \binom 65 = 36.$
There is no way for you to get $9$ correct. You get $8$ correct if and only if the correct answers are "yes" on $4$ of your "yes" answers and "no" on your other "yes" answers. The number of ways this can happen is the number of ways to choose $4$ correct answers from your $6$ "yes" answers, times the number of ways to choose $4$ correct answers from your $6$ "no" answers: $\binom 64 \times \binom 64 = 225.$
In any other case you fail. So the chance to pass is $$ \frac{1 + 36 + 225}{924} = \frac{131}{462} \approx 0.283550, $$ which is much better than the chance of passing if you simply toss a coin for each individual question but not nearly as good as getting $4$ or more heads in $6$ coin tosses.
Just to check, we can compute the chance of failing in the same way: $6$ answers correct ($3$ "yes" and $3$ "no"), $4$ answers correct, $2$ correct, $0$ correct. This probability comes to $$ \frac{\binom 63^2 + \binom 62^2 + \binom 61^2 + 1}{924} = \frac{400 + 225 + 36 + 1}{924} = \frac{331}{462} \approx 0.716450, $$ which is the value needed to confirm the answer above.
In terms of balls and urns: Maybe it helps to think about it as follows:
You have a red urn and a blue urn, and you have $6$ red balls and $6$ blue balls. You randomly put $6$ of the twelve balls in the red urn, and the other $6$ in the blue urn. Now: what it the chance that at least $8$ balls are in the 'right' (i.e. same colored) urn?
Well, to get $8$ correct, you either need to get all $6$ red balls in the red urn ($1$ possibility), or $5$ red ones and $1$ blue in the red urn (${6 \choose 5} \cdot {6 \choose 1} = 6 \cdot 6 = 36$ possibilities), or $4$ red ones and $2$ blue ones (${6 \choose 4} \cdot {6 \choose 2} = 15 \cdot 15 = 225$ possibilities). This is out of a total of ${12 \choose 6} = 924$ possibilities, and so the probability is $\frac{1+36+225}{924}$
NOTE: Thanks to @DavidK for pointing out my initial answer was wrong! Everyone please upvote his answer!
Try the following approach.
First, assume that the answerer puts "yes" for the first six and "no" for the last six. Since the order of his answers clearly cannot change his probability of passing, this is an okay assumption.
Note that if there are $x$ questions correct on the first half of the test, there are $x$ questions correct on the second half as well (think about why). So it's sufficient to get the probability that $4$ of the first $6$ answers are yes.
Now, try to calculate the probability that a randomly ordered row of $12$ balls -- $6$ "yes" balls and $6$ "no" balls -- has $4$ "yes" balls among the first $6$ balls in the row.