Probability Question 185
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
Options:
A) $ \frac{37}{40} $
B) $ \frac{1}{37} $
C) $ \frac{36}{37} $
D) $ \frac{1}{9} $
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Answer:
Correct Answer: B
Solution:
We define the following events :
$ A_1: $ He knows the answer.
$ A_2: $ He does not know the answer.
$ E: $ He gets the correct answer. Then $ P(A_1)=\frac{9}{10},P(A_2)=1-\frac{9}{10}=\frac{1}{10},P( \frac{E}{A_1} )=1, $
$ P( \frac{E}{A_2} )=\frac{1}{4} $
$ \therefore $ Required probability $ =P( \frac{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{1}{37}. $
BETA
