# Exam with 1212 yes/no questions (half yes, half no) and 88 correct needed to pass, is it better to answer randomly or answer exactly 6 times yes?

In an exam with $$1212$$ yes/no questions with $$88$$ correct needed to pass, is it better to answer randomly or answer exactly $$66$$ times yes and 6 times no, given that the answer ‘yes’ is correct for exactly $$66$$ questions?

I have calculated the probability of passing by guessing randomly and it is

$$12∑k=8(12k)0.5k0.5n−k=0.194\sum_{k=8}^{12} {{12}\choose{k}}0.5^k0.5^{n-k}=0.194$$

Now given that the answer ‘yes’ is right exactly $$66$$ times, is it better to guess ‘yes’ and ‘no’ $$66$$ times each?

My idea is that it can be modelled by drawing balls without replacement. The balls we draw are the correct answers to the questions.

Looking at the first question, we still know that there are $$66$$ yes and no’s that are correct. The chance that a yes is right is $$612\frac{6}{12}$$ and the chance that a no is right is also $$612\frac{6}{12}$$.

Of course the probability in the next question depends on what the first right answer was. If yes was right, yes will be right with a probability of $$5/115/11$$ and a no is right with the chance $$6/116/11$$. If no was right, the probabilities would change places.

Now that we have to make the choice $$1212$$ times and make the distinction which one was right, we get $$2122^{12}$$ paths total. We cannot know what the correct answers to the previous questions were. So we are drawing $$1212$$ balls at once, but from what urn? It cannot contain $$2424$$ balls with $$1212$$ yes and $$1212$$ no’s. Is this model even correct?

Is there a more elegant way to approach that?

I am asking for hints, not solutions, as I’m feeling stuck. Thank you.

Edit: After giving @David K’s answer more thought, I noticed that the question can be described by the hypergeometric distribution, which yields the desired result.

## Answer

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

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

which is the value needed to confirm the answer above.

Attribution
Source : Link , Question Author : B.Swan , Answer Author : David K