Solve the following question and mark the best possible option.
In a test, an examinee either guesses or copies or knows the answers to a multiple-choice question with four choices. The probability that he makes a guess is `1/3` and the probability that he copies the answer is `1/6`. The probability that his answer is correct given that he copied it, is `1/8`. Find the probability that he knew the answer to the question given that he correctly answered it.
Let E1 be the event that the answer is guessed E2 be the event that the answer is copied, E3 be the event that the examinee knows the answer and E be the event that the examinee answered correctly.
Given P(E1) = `1/3, P(E_2) = 1/6`
Assume that events E1, E2 and E3 are exhaustive.
P(E1) + P(E2) + P(E3) = 1
P(E3) = 1 - P(E1) - P(E2) = 1 - `1/3 - 1/6 = 1/2`
Now P(E/E1) = Probability of getting a correct answer by guessing = `1/4`
[Since these are 4 alternatives]
P(E/E2) = Probability of answering correctly by copying = `1/8`
P(E/E3) = Probability of answering correctly by knowing = 1
Clearly (E3/E) is the event he knew the answer to the question given that he correctly answered it.
∴ P(E3/E) = `[P(E_3) P(E/E_3)]/[P(E_1) P(E/E_1) + P(E_2) P(E/E_2) + P(E_3) P(E/E_3)]`
∴ P(E3/E) = `(1/2 xx 1)/(1/3 xx 1/4 + 1/6 xx 1/8 + 1/2 xx 1) = 24/29.`