I am a little bit confused about the following math problem I asked myself. Let's say there are 10 exams in a year, and each exam contains exactly three questions. A question's probability to be answered correctly by a student is given by 1/4. (If a question is not answered at all, we say it is not answered correctly.) Now assume that after each exam the teacher chooses one out of the three tasks to be corrected, so the other two will be ignored. Each task has the same probability to be choosen by the teacher. If a student gets the chosen question right, he gets one point.
My question is: What is the probability of a student to have a least 5 points in total at the end of the year?
I thought about using Binomial distribution, but the problem is that my question is a "choose with replacement" and you do a choose without replacement when you use Binomial dsitribution...
Any help is appreciated :) Thanks!
"Choose with replacement" refers to modeling a process where the outcome of one trial does not influence the outcomes of any other trials. So for instance, when we flip a coin, the outcome of subsequent flips does not depend on the outcome of previous flips. Or, if we draw balls from an urn with balls of two different colors, if we replace the ball after each draw, then the outcomes of future draws does not depend on the outcomes of previous draws--unlike the case where we do not replace the ball.
As this notion applies to your specific question about exams, the intent is that each exam is an independent trial with its own outcome, so we are dealing with a binomial model because the performance on future exams does not depend on the performance on previous exams. Whether that is a realistic assumption is a separate matter--this is what we are told to assume when it is stipulated that the student as a $1/4$ chance to get any single question correct.
Another way to think about it is that the exams are not reused--each one is different, thus in a sense the "replacement" that is going on is the replacement of one exam with another.
Note that, for all intents and purposes, the professor could simply choose to grade the first of the three questions in each exam, because the outcome for each question has the same probability distribution as any of the other questions. And then, the second and third questions become irrelevant information, so our model for the number of correct questions is simply binomial with $n = 10$ and $p = 1/4$.
However, things get more interesting if, for example, within a single test, the questions have increasing difficulty, so although they remain independent, they are not identically distributed: say, the chance of the first question being answered correctly is $3/4$, the second $1/2$, and the third and most difficult has only $1/4$ chance of being answered correctly. Then the professor's random choice of which question to grade becomes relevant.