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The problem forumaltes as:

Student takes test of 12 questions, each question is single-choice with 4 options. For each question he answers correctly if he knows the question, otherwise he guesses uniformly. Assume student answered correctly for exactly 9 questions. What is probability that he knew at least 7 questions? My question is the following: does this problem statement make sense unless we have a priori probability distribution of how many questions does student know?

If it makes sense, how to solve it, and if it doesn't, how to show that and what could be the ways to complete the statement of this problem so that it would make sense?

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  • $\begingroup$ This question does make sense, assuming the Bayesian conditional probability model. It might help to read up on how it works. $\endgroup$ Commented Apr 6, 2019 at 16:50
  • $\begingroup$ I think this makes sense right away. Define events $A_k = $ "the student knows $k$ questions", $B_k$ = "the student answers $k$ questions right" . You are looking for $\mathbb{P}(A_1 \cup \dots \cup A_7 | B_9)$ and you can use Bayes formulas to compute this. $\endgroup$ Commented Apr 6, 2019 at 16:52
  • $\begingroup$ @Gâteau-Gallois am I wrong that if we use direct bayes formula, we will need to know $P(A_i)$ which is not given? $\endgroup$ Commented Apr 6, 2019 at 17:00

1 Answer 1

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Let $k$ denote the number of questions the students genuinely knows how to answer. Clearly $k\le 9$. Then the probability of scoring $9$ is $\frac{\binom{12-k}{3}3^3}{4^{12-k}}$. Using Bayes's theorem, we can write $k$'s probability distribution conditional on a score $S=9$ as $$P(k=r|S=9)=\frac{\binom{12-r}{3}3^3}{4^{12-r}}\frac{P(k=r)}{P(S=9)}.$$In particular,$$P(k=7|S=9)=\frac{\binom{5}{3}3^3}{4^5}\frac{P(k=7)}{P(S=9)}.$$But $P(k=7)$ is a Bayesian prior, and $P(S=9)$ is obtained from that prior. So if you want to get a specific numerical value for the probability, we need more information. For example, suppose the questions are independently known with probability $p$, so $$P(k=r)=\binom{12}{r}p^r(1-p)^{12-r},\,P(S=9)=\sum_{r=0}^9\frac{\binom{12-r}{3}3^3}{4^{12-r}}P(k=r).$$Ultimately, $P(k=7|S=9)$ will depend on $p$. Bayesian statistics always starts from a prior; the posterior depends on the prior, but with less sensitivity to it given a large dataset.

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