# In an entrance test there are multiple choice questions. There are four possible answers to each question of which one is correct.

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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) $\frac{1}{9}$

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Answer: (b) $\frac{1}{37}$

We define the given events as:

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

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

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