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.

#### Options

`13/27`

`24/29`

`18/31`

`17/31`

#### Solution

Let E_{1} be the event that the answer is guessed E_{2} be the event that the answer is copied, E_{3 }be the event that the examinee knows the answer and E be the event that the examinee answered correctly.

Given P(E_{1}) = `1/3, P(E_2) = 1/6`

Assume that events E_{1}, E_{2 }and E_{3} are exhaustive.

P(E_{1}) + P(E_{2}) + P(E_{3}) = 1

P(E_{3}) = 1 - P(E_{1}) - P(E_{2}) = 1 - `1/3 - 1/6 = 1/2`

Now P(E/E_{1}) = Probability of getting a correct answer by guessing = `1/4`

[Since these are 4 alternatives]

P(E/E_{2}) = Probability of answering correctly by copying = `1/8`

P(E/E_{3}) = Probability of answering correctly by knowing = 1

Clearly (E_{3}/E) is the event he knew the answer to the question given that he correctly answered it.

∴ P(E_{3}/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(E_{3}/E) = `(1/2 xx 1)/(1/3 xx 1/4 + 1/6 xx 1/8 + 1/2 xx 1) = 24/29.`