When we have a midterm exam of a probability course in a classroom, we often design two versions of the midterm exam paper. We do not have cameras in the classroom. In November, 2019, each version had 10 questions. Version A was the same as Version B except for two questions: in each of those two questions, two versions were different only on one number. Difference Version A Version B Question 2 P(A)30.4 Question 7 a3D0.05 P(A)-0.5 a=0.10 Student John worked on Version A, but his solutions were using the numbers of Version B. The student at John's left side and the student at John's right side had Version B. As we know well, it is possible for a pérson to make mistakes on reading numbers. According to a research report in 1950, the probability that a person had this type of mistakes in one question was less than 1%. Suppose that a person makes the mistake in different questions are independent. The probability that John did the exam by himself would be less than 0.01% from 0.012-0.01%. Hence, the instructor claimed that John cheated. However, John did not cheat. Now, John is going to appeal to the fair committee. How can John argue that he did not cheat?

A First Course in Probability (10th Edition)
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ISBN:9780134753119
Author:Sheldon Ross
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Chapter1: Combinatorial Analysis
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Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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When we have a midterm exam of a probability course in a classroom, we often design two versions of the
midterm exam paper. We do not have cameras in the classroom.
In November, 2019, each version had 10 questions. Version A was the same as Version B except for two
questions: in each of those two questions, two versions were different only on one number.
Difference Version A Version B
Question 2 P(A)3D0.4
Question 7 a=0.05
P(A)-0.5
a=0.10
Student John worked on Version A, but his solutions were using the numbers of Version B. The student at
John's left side and the student at John's right side had Version B.
As we know well, it is possible for a person to make mistakes on reading numbers. According to a research
report in 1950, the probability that a person had this type of mistakes in one question was less than 1%.
Suppose that a person makes the mistake in different questions are independent. The probability that John did
the exam by himself would be less than 0.01% from 0.012-0.01%. Hence, the instructor claimed that John
cheated.
However, John did not cheat. Now, John is going to appeal to the fair committee. How can John argue that he
did not cheat?
Transcribed Image Text:When we have a midterm exam of a probability course in a classroom, we often design two versions of the midterm exam paper. We do not have cameras in the classroom. In November, 2019, each version had 10 questions. Version A was the same as Version B except for two questions: in each of those two questions, two versions were different only on one number. Difference Version A Version B Question 2 P(A)3D0.4 Question 7 a=0.05 P(A)-0.5 a=0.10 Student John worked on Version A, but his solutions were using the numbers of Version B. The student at John's left side and the student at John's right side had Version B. As we know well, it is possible for a person to make mistakes on reading numbers. According to a research report in 1950, the probability that a person had this type of mistakes in one question was less than 1%. Suppose that a person makes the mistake in different questions are independent. The probability that John did the exam by himself would be less than 0.01% from 0.012-0.01%. Hence, the instructor claimed that John cheated. However, John did not cheat. Now, John is going to appeal to the fair committee. How can John argue that he did not cheat?
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