These days competitive examinations are held in online mode. During the examination student either guesses or copies or knows the answer of a multiple choice question (MCQ) with four choices each. The probability that a student makes guesses is \(\displaystyle\frac{1}{3}\) and that he/she copies \(\displaystyle\frac{1}{6}\). The probability of marking answer correctly when he/she copied is \(\displaystyle\frac{1}{8}\) and probability of making answer correctly when he/she guesses is \(\displaystyle\frac{1}{4}\). It is also assumed that a student marked answer correctly if he/she knows the answer.
By using the above mentioned information answer following question.
If answer is correct, the probability that he guesses, is :
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{4}{29}\)
Let \(E_1\), \(E_2\), \(E_3\) be the events that the student guesses, copies and knows the answer, and let A be the event that the answer is correct.
\( \displaystyle P(E_1) = \frac{1}{3},\quad P(E_2) = \frac{1}{6},\quad P(E_3) = 1 - \frac{1}{3} - \frac{1}{6} = \frac{1}{2} \)
\( \displaystyle P(A|E_1) = \frac{1}{4},\quad P(A|E_2) = \frac{1}{8},\quad P(A|E_3) = 1 \)
By the theorem of total probability,
\( \displaystyle P(A) = \frac{1}{3} \times \frac{1}{4} + \frac{1}{6} \times \frac{1}{8} + \frac{1}{2} \times 1 \)
\( \displaystyle = \frac{4}{48} + \frac{1}{48} + \frac{24}{48} = \frac{29}{48} \)
By Bayes’ theorem,
\( \displaystyle P(E_1|A) = \frac{P(E_1)\,P(A|E_1)}{P(A)} = \frac{4/48}{29/48} \)
\( \displaystyle = \frac{4}{29} \)
Note that \(\frac{1}{4}\) is P(correct | guesses), the probability the other way round.