(apply total probability and Bayes\' rules) In a recent survey in a Statistics c
ID: 3044031 • Letter: #
Question
(apply total probability and Bayes' rules) In a recent survey in a Statistics class, it was determined that only 60% of the students attend class on Thursday. From past data it was noted that 98% of those who went to class on Thursday pass the course, while only 20% of those who did not go to class on Thursday passed the course. a) What percentage of students is expected to pass the course? (15 points) b) Given that a student passes the course, what is the probability that he/she attended classes on Thursday? (15 points)Explanation / Answer
A)
Let A be the event of passing and B be the event of attending on Fridays.
Given
P(A|B) = 0.98 P(A|B') = 0.2
P(B) = 0.6 => P(B') = 1 - 0.6 = 0.4
P(A B) = P(A|B) P(B)
= 0.98 * 0.6
= 0.588
P(A B') = P(A|B') P(B)
= 0.2 * 0.4
= 0.08
By total probability
P(A) = P(A B) + P(A B')
= 0.588 + 0.08
= 0.668.
b.)
By bayes' rule
P(B|A) = P(A B) / P(A)
= 0.588 / 0.668
= 0.8802.
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