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Detecting Cheating in Examination

XAT 2013 · Decision Making · Medium · Decision Making

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Passage / data set

A teacher wanted to administer a multiple choice (each question having six choices) based quiz of high difficulty level to a class of sixty students. The quiz had sixty questions. The probability of selecting the correct answer for a good student and a brilliant student was 0.2 and 0.25 respectively. The poor students had no learning advantage. The teacher did not want students to cheat but does not have time and resources to monitor. All students were seated serially in 10 rows and 6 columns.

Question 1 of 2

Is it possible for the teacher to detect cheating without monitoring? Choose the statement that best describes your opinion:

  1. A.

    It is not at all possible; teacher will have to introduce technology if there is no human support.

  2. B.

    It is always possible; but teacher has to calculate the exact answer.

  3. C.

    It is possible when many students sitting next to each other have the same incorrect answers for multiple questions. However, there can be a small error in judgment.

  4. D.

    It is possible when many students sitting next to each other have the same correct answers for multiple questions. However, there can be a small error in judgment.

  5. E.

    It is possible only for poor students but not for good and brilliant students. However, there can be a small error in judgment.

Answer

C

Explanation

With 6 options, the probability of guessing the same wrong option by chance across multiple questions is extremely low. Thus, identical incorrect answers among neighbors indicate cheating.

Question 2 of 2

Three good students were seated next to each other. What is the probability of them having the same incorrect choice for four consecutive questions?

  1. A.

    256/390625

  2. B.

    256/3125

  3. C.

    4/3125

  4. D.

    1/3125

  5. E.

    Cannot be calculated

Answer

E

Explanation

The problem does not specify the probability distribution of choosing among the 5 wrong choices for each student, so exact probability cannot be calculated without assuming uniform distribution.

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