Probability theory a problem about students

Question

There are M questions on an exam. John knows answer to N of them. When does he have bigger probability of passing the exam:

1) Will he be the first to take the exam
2) He will take the exam at some place in the queue
(After passing the exam, the student picks up their ticket)

For the first situation P(A) = N/M If he is second in line $$P(A) = P(H_1)P(A|H_1)+P(H_2)P(A|H_2) = \\ = \frac{N}{M} \frac{N-1}{M-1} + \frac{M-N}{M} \frac{N}{M-1} = \frac{N}{M}\left( \frac{N-1}{M-1} + \frac{M-N}{M-1}\right) = \frac{N}{M}$$ But how can we prove that the probability does not depend on the position for $k>2$ ?

Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. — José Carlos Santos, Mar 16 '20 at 07:51
I don't understand what "the student picks up their ticket" means. What information is on this ticket? Do all the students show their ticket to John? — Gerry Myerson, Mar 16 '20 at 09:51
Each ticket contains one question from the exam program. John does not know what tickets students will receive. It is necessary to prove that the probability does not depend on where John is (First, second, third ...) — Олег Кутузов, Mar 16 '20 at 11:11
But what do the tickets have to do with anything? How is John affected by which student receives which ticket? — Gerry Myerson, Mar 16 '20 at 12:19
I'm voting to close because I can't understand what you're asking. — Gerry Myerson, Mar 16 '20 at 12:21
He has learned only N questions from M. If he gets the exam ticket that he has learned, he will pass the exam. If not, it will fail. Students ahead of him take their tickets and, accordingly, these questions will no longer be. — Олег Кутузов, Mar 16 '20 at 12:55
Does this answer your question? If you draw two cards, what is the probability that the second card is a queen? — JMoravitz, Mar 16 '20 at 13:57
You should edit that clarification into the body of your question, Oner, not make people go through all these comments to understand it. Each "ticket" has only one question? Each student is asked to answer just one question? — Gerry Myerson, Mar 18 '20 at 00:25

score 2 · Answer 1 · answered Mar 16 '20 at 12:07

Your statement of the problem is very unclear. If I have decoced it correctly, each student is asked a different one of the $M$ questions and passes the exam if she answers this one question correctly.

The fact that John’s chances of passing the exam don’t depend on his place in the queue is much easier to prove by symmetry than with the sort of calculation you carried out for the case that John is in second place. There is nothing to distinguish any question from any other question. Thus, irrespective of his place in the queue, John must have the same probability to get any of the questions as any other, by symmetry alone.

Probability theory a problem about students

1 Answers1