In an entrance test there are multiple choice questions. There are four possible answers to each questions, of which one is correct. The probability that a student knows the answer to a question is 9/10. If he gets the correct answer to a question, then the probability that he was guessing is

$\frac{37}{40}$

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$\frac{1}{37}$

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$\frac{36}{37}$

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$\frac{1}{9}$

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Solution

The correct option is B
$\frac{1}{37}$

$E_{1} \to knows the answer E_{2} \to guesses the answer E \to gets correct answer P (E_{1}) = \frac{9}{10}, P (\frac{E}{E_{1}}) = 1 P (E_{2}) = \frac{1}{10}, P (\frac{E}{E_{2}}) = \frac{1}{4} P (\frac{E_{2}}{E}) = \frac{1}{37}$

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In an entrance test there are multiple choice questions. There are four possible answers to each questions, of which one is correct. The probability that a student knows the answer to a question is 9/10. If he gets the correct answer to a question, then the probability that he was guessing is

The correct option is B 137 E1→ knows the answerE2→ guesses the answerE→ gets correct answerP(E1)=910 , P(EE1)=1P(E2)=110 , P(EE2)=14P(E2E)=137

The correct option is B
$\frac{1}{37}$

$E_{1} \to knows the answer E_{2} \to guesses the answer E \to gets correct answer P (E_{1}) = \frac{9}{10}, P (\frac{E}{E_{1}}) = 1 P (E_{2}) = \frac{1}{10}, P (\frac{E}{E_{2}}) = \frac{1}{4} P (\frac{E_{2}}{E}) = \frac{1}{37}$