Probability Question 274
Question: In a test, an examinee either guesses or copies or knows the answer to a multiple choice question with four choices. The probability that he makes a guess is $ \frac{1}{3}. $ The probability that he copies is $ \frac{1}{6} $ and the probability that his answer is correct given that he copied it is $ \frac{1}{8}. $ The probability that he knew the answer to the question given that he correctly answered it, is
Options:
A) $ \frac{24}{29} $
B) $ \frac{1}{4} $
C) $ \frac{3}{4} $
D) $ \frac{1}{2} $
Show Answer
Answer:
Correct Answer: A
Solution:
Let $ E_1 $ be the event that the answer is guessed, $ E_2 $ be the event that the answer is copied, $ E_2 $ be the event that the examinee knows the answer end E be the event that the examinee answer correctly. Given $ P(E_1)=\frac{1}{3},P(E_2)=\frac{1}{6} $ Assume that events $ E_1,E_2\And E_3 $ are exhaustive.
$ \therefore P(E_1)+P(E_2)+P(E_3)=1 $
$ \therefore P(E_3)=1-P(E_1)-P(E_2)=1-\frac{1}{3}-\frac{1}{6}=\frac{1}{2} $ Now, $ P( \frac{E}{E_1} )= $ Probability of getting correct answer by guessing $ =\frac{1}{4} $ (Since 4 alternatives) $ P( \frac{E}{E_2} ) $ = Probability of answering correctly by copying $ =\frac{1}{8} $ And $ P( \frac{E}{E_3} )= $ Probability of answering correctly be knowing = 1 Clearly, $ ( \frac{E_3}{E} ) $ is the event he knew the answer to the question given that he correctly answered it. Using Baye’s theorem $ P( \frac{E_3}{E} ) $ $ =\frac{P(E_3).P( \frac{E}{E_3} )}{P(E_1).P( \frac{E}{E_1} )+P(E_2).P( \frac{E}{E_2} )+P(E_3).P( \frac{E}{E_3} )} $ $ =\frac{\frac{1}{2}\times 1}{\frac{1}{3}\times \frac{1}{4}+\frac{1}{6}\times \frac{1}{8}+\frac{1}{2}\times }=\frac{24}{29} $
BETA
