In an entrance test there are multiple choice questions. There are fou

1.	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» 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 \(P(A_2/E) = \frac{P(A_2).P(E/A_2)}{P(A_1).P(E/A_1)+P(A_2).P(E/A_2)}\) = \(\frac{\frac{1}{10}.\frac{1}{4}}{\frac{9}{10}.1+\frac{1}{10}.\frac{1}{4}}\) = \(\frac{\frac{1}{40}}{\frac{37}{40}}\) = \(\frac{1}{37}\)

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»

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

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

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