Combinatorics question students Bayes theorem

Question

I have this task and I would like to hear ideas/solution for it.

There are 30 students on an exam. The questions are 20.

Every student draws 2 question and the condition to pass the exam is true answer on the both questions.

There are 3 groups of students:

1. 5 students knows all questions $ (\dfrac{5}{30}) $

2. 15 students knows 10 questions $ (\dfrac{15}{30} ) $

3. 10 students knows 5 questions $ (\dfrac{10}{30} )$

Question #1: What is the probability random student to pass the exam?

Questions #2: If we know, that 1 student had passed the exam, what is the probability that the student knew only 5 questions?

My solution: #1 I'm summing the probability of every student group $[\dfrac{5}{30}\cdot1\cdot1] + [\dfrac{15}{30} \cdot \dfrac{10}{20}\cdot\dfrac{10}{20}] + [ \dfrac{10}{30} \cdot \dfrac{5}{20} \cdot \dfrac{5}{20}] = 0.3053$

#2 $\dfrac{10}{30} \cdot \dfrac{5}{20} \cdot \dfrac{4}{19} = 0.017; $

$ 0.3053 \cdot 0.017 = 0.0051$

I don't think my fractions are right..

Answer 1

Take for example the ten students who know only five of the twenty questions. In order for a student in this group to pass the exam, both questions they are given must be among the five that they know how to answer. There are $\binom{5}{2} = 10$ ways to choose two questions from the twenty such that both come from the group of five that they know how to answer, but there are $\binom{20}{2} = 190$ ways to choose two questions regardless of whether the student knows the answer. Therefore, the probability that a student in this group will pass is $10/190 = 1/19$.

If we apply similar reasoning to the students who know $10$ out of the $20$ questions, we see that their passing probability is $$\frac{\binom{10}{2}}{\binom{20}{2}} = \frac{45}{190} = \frac{9}{38}.$$

And of course, the students who know all questions will always pass; their probability is $1$.

Now we weight these individual passing probabilities by the probability of picking a student from the respective category to get an overall probability of passing for a randomly selected student:

$$1 \cdot \frac{5}{30} + \frac{9}{38} \cdot \frac{15}{30} + \frac{1}{19} \cdot \frac{10}{30} = \frac{23}{76}.$$

For the second part, you are told that the randomly selected student passed the exam. This affects the posterior probability of that student having belonged to the group that knew only five questions. Intuition suggests this probability is extremely small.

We use Bayes theorem to calculate this. Let $P$ be the event a randomly selected student passed. Let $F$ be the event a randomly selected student belongs to the group that knows only five questions. Then we are asked to compute $\Pr[F \mid P]$. By Bayes theorem, this is $$\Pr[F \mid P] = \frac{\Pr[P \mid F]\Pr[F]}{\Pr[P]}.$$ We are given $\Pr[F] = \frac{10}{30}$. We already calculated in the first part $$\Pr[P \mid F] = \frac{1}{19}.$$ And we calculated $$\Pr[P] = \frac{23}{76}.$$ So all that is left is to substitute these values and simplify.

---

Now, the idea that an exam consists of only two questions, and both must be answered correctly to pass, seems a little unrealistic. So maybe we can consider a modification of the problem. Suppose instead that the exam consists of $4$ questions, and in order to pass, a student must answer at least $3$ of the $4$ questions correctly. As before, however, the number of questions that a student knows how to answer is the same.

Then for instance, among the group of students who know only five of the twenty questions, the probability of passing is $$\Pr[P \mid F] = \frac{\binom{5}{3} \binom{15}{1}}{\binom{20}{4}} + \frac{\binom{5}{4}\binom{15}{0}}{\binom{20}{4}}.$$ This probability arises from a hypergeometric distributon.

How would you compute the probabilities for the other two groups?

Answer 2

For 1 $$P(\text{student passes}) = \frac{5}{30} \cdot 1 + \frac{15}{30}\frac{\binom{10}{2}}{\binom{20}{2}} + \frac{10}{30}\frac{\binom{5}{2}}{\binom{20}{2}} $$ splitting on the type of student and if we know e.g. 10 questions, we have to pick 2 from the good questions over 2 of all 20 questions, hence the binomials. Your squares treat it like the same question could be chosen twice.

This simplifies to $\frac{69}{228}$ which is about $0.3026$.

Q. 2 asks for $P(\text{type3 student}| \text{ student passes})$ which is $$\frac{P(\text{type3 student and student passes})}{P(\text{ student passes})}$$

by the definition of conditional probability. In 1 we already found the denominator, the numerator equals $\frac{10}{30}\frac{\binom{5}{2}}{\binom{20}{2}} =\frac{4}{228}$. So we get $\frac{4}{69}$ as the answer.