MHTCET 18 Apr 2026 Shift 2 Paper

Concept:Use Bayes' theorem to find the conditional probability that the student was guessing, given that the answer is correct.Explanation:Let $K$ be the event that the student knows the answer.Let $K'$ be the event that the student does not know the answer, so he guesses.Let $C$ be the event that the answer is correct.Given probability that the student knows the answer:$$P(K) = 90\% = \frac{9}{10}$$Therefore,$$P(K') = 1 - \frac{9}{10} = \frac{1}{10}$$If the student knows the answer, he definitely answers correctly:$$P(C \mid K) = 1$$If he guesses among 4 options with only one correct answer:$$P(C \mid K') = \frac{1}{4}$$We need the probability that he was guessing, given that he answered correctly:$$P(K' \mid C)$$Using Bayes' theorem:$$P(K' \mid C) = \frac{P(K') \cdot P(C \mid K')}{P(K) \cdot P(C \mid K) + P(K') \cdot P(C \mid K')}$$Substitute the values:$$P(K' \mid C) = \frac{\frac{1}{10} \times \frac{1}{4}}{\frac{9}{10} \times 1 + \frac{1}{10} \times \frac{1}{4}}$$$$= \frac{\frac{1}{40}}{\frac{9}{10} + \frac{1}{40}}$$$$= \frac{\frac{1}{40}}{\frac{36}{40} + \frac{1}{40}}$$$$= \frac{\frac{1}{40}}{\frac{37}{40}}$$$$= \frac{1}{37}$$Answer:The required probability is $\frac{1}{37}$.Correct option: B.