Consider the case of the girl who was accused of cheating on a multiple choice t
ID: 3172727 • Letter: C
Question
Consider the case of the girl who was accused of cheating on a multiple choice test. On one of the questions, the numbers of students out of 88 choosing answers A,B,C,D, and E were as follows. A B C D E
10 12 32 11 23
We estimate the probability that a given student would select a given answer by the proportion of the class that selected that answer. For example, the probability that a given student chooses answer A is estimated to be 10/88.
a.What is the estimated probability that she and her neighbor would get the same answer, assuming there was no cheating?
b..What is the estimated probability that she and her neighbor would get the same answer, given that her neighbor selected answer B, assuming there was no cheating?
Explanation / Answer
Solution:-
10 12 32 11 23
a) The estimated probability that she and her neighbor would get the same answer, assuming there was no cheating is 0.2477.
P(x, x) = P(A, A) + P(B, B) + P(C, C) + P(D, D) +P(E, E)
P(x, x) = 0.0129 + 0.018595 + 0.1322314 + 0.015625 + 0.06831
P(x, x) = 0.2477
b. The estimated probability that she and her neighbor would get the same answer, given that her neighbor selected answer B, assuming there was no cheating is 0.1364
Number of students marking B = 12
Total students = 88
The probability of getting B for her = 12/88 = 0.1364