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?
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.