Question

There are 3 True or False questions in an exam, if a candidate knows the answer she/he answers it correctly, otherwise, a guess is made and the probability of getting it right is 1/2. An examiner assumes that every candidate knows no answer, 1 answer, 2 answers, 3 answers with equal probabilities, a candidate answered two of the three questions correctly. What is the probability that this candidate knew the answer to only one of them?

(A) 1/11.

(B) 2/11.

(C) 3/11.

(D) 4/11.

Answer 1

P(knows no answer) = 0.25

P(knows 1 answers) = 0.25

P(knows 2 answers) = 0.25

P(knows 3 answers) = 0.25

Bayes' Theorem: P(A | B) = P(A & B)/P(B)

P(candidate knew only one of the correct answer | candidate answered 2 of the three questions correctly) = P(candidate knew only one of the correct answer and answered 2 of the three questions correctly) / P(candidate answered 2 of the three questions correctly)

= P(candidate knew only one of the correct answer and answered 2 of the three questions correctly) / [P(candidate knew none of the correct answer and answered 2 of the three questions correctly) + P(candidate knew one of the correct answer and answered 2 of the three questions correctly) + P(P(candidate knew two of the correct answer and answered 2 of the three questions correctly)]

= [P(knows 1 correct answer and gueses 1 of the other 2 correctly)]/[P(knows 0 correct answers and gueses 2 of the 3 correctly) + P(knows 1 correct answer and gueses 1 of the other 2 correctly) + P(knows 2 correct answer and gueses the other one incorrectly)]

= (0.25 x 2x0.5²) / (0.25 x 3C2x0.5³ + 0.25 x 2 x 0.5² + 0.25x0.5)

= (0.25x0.5)/(0.25 x 0.375 + 0.25x0.5 + 0.25x0.5)

= 4/11