Question

In an entrance test there are multiple choice questions. There are four possible answers to each question of which one is correct.The probability that a student knows the answer to a question is 90%. If he gets the correct answer to a question, then the probability that he was guessing is 

(a) \(\frac{37}{40}\)

(b) \(\frac{1}{37}\)

(c) \(\frac{36}{37}\)

(d) \(\frac{1}{9}\)

Answer 1

Answer: (b) \(\frac{1}{37}\)

We define the given events as: 

A1: Student knows the answer

A2: Student does not know the answer 

E: He gets the correct answer.

P(A₁) =\(\frac{9}{10}\) , P(A₂) = 1-\(\frac{9}{10}\)  = \(​​\frac{1}{10}\)

\(\therefore\)   P(E/A1) = P(Student gets the correct answer when he knows the answer) = 1 

P(E/A2) = P(Student gets the correct answer when he does not know the correct answer) = 1/4 

\(\therefore\)  Required probability 

\(P(A_2/E) = \frac{P(A_2).P(E/A_2)}{P(A_1).P(E/A_1)+P(A_2).P(E/A_2)}\)

= \(\frac{\frac{1}{10}.\frac{1}{4}}{\frac{9}{10}.1+\frac{1}{10}.\frac{1}{4}}\) = \(\frac{\frac{1}{40}}{\frac{37}{40}}\) = \(\frac{1}{37}\)