Question

Consider an automated plagiarism detection software that is used to evaluate essay submissions. Four sections of a writing course use the software to check for plagarism, with 30% of the students in section 1, 16% in section 2, 30% in section 3, and 24% in section 4. In section 1 of a course, 20% of the essays are flagged, in section 2, 23%, section 3, 15% and section 4, 8%. (a) What percentage of total students committed plagiarism overall? (b) Given that a particular student committed plagiarism, what in the probability that they were registered for section 1 of the course. (c) Given that a particular student committed plagiarism, what in the probability that they were registered for section 2 of the course. (d) If there are 200 students registered between these 4 sections, how many students in section 3 cheated?

Answer 1

We are given here that:  Four sections of a writing course use the software to check for plagarism, with 30% of the students in section 1, 16% in section 2, 30% in section 3, and 24% in section 4, therefore we have here:
P( section 1) = 0.3,
P( section 2) = 0.16
P( section 3) = 0.3
P( section 4) = 0.24

Also, we are given here that: In section 1 of a course, 20% of the essays are flagged, in section 2, 23%, section 3, 15% and section 4, 8%. Therefore, we have here:
P( flagged | section 1) = 0.2
P( flagged | section 2) = 0.23
P( flagged | section 3) = 0.15
P( flagged | section 4) = 0.08

a) Using law of total probability, we get here:
P( flagged ) = P( flagged | section 1) P(section 1) + P( flagged | section 2) P(section 2) + P( flagged | section 3) P(section 3) + P( flagged | section 4) P(section 4)
P( flagged ) = 0.2*0.3 + 0.23*0.16 + 0.15*0.3 + 0.08*0.24 = 0.161

Therefore 16.1% of the total students committed plagiarism overall.

b)  Given that a particular student committed plagiarism, the probability that they were registered for section 1 of the course is computed using bayes theorem as:
P( section 1 | flagged ) = P( flagged | section 1) P(section 1) / P( flagged )
P( section 1 | flagged ) = 0.2*0.3 / 0.161 = 0.3727

Therefore 0.3727 is the required probability here.

c) Given that a particular student committed plagiarism, the probability that they were registered for section 2 of the course is computed using bayes theorem as:
P( section 2 | flagged ) = P( flagged | section 2) P(section 2) / P( flagged )
P( section 2 | flagged ) = 0.23*0.16 / 0.161 = 0.2286

Therefore 0.2286 is the required probability here.

d) If there are 200 students registered between these 4 sections, number of students in section 3 who cheated is computed here as:

= 200*P( section 3) P( flagged | section 3) = 200*0.15*0.3 = 9

Therefore 9 are the number of students from section 3 who cheated.