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 make a guess is 13 and the probability that he copies the answer is 16. The probability that his answer is correct given that he copied it is 18 then the probability that he know the answer to the questions given that he correctly answered it is

E1 : gauss the answer. E2 : Copies the answer E3 : Known the answer, E : answer is correctP(E1)=13,P(E2)=16,P(E3)=1−(13+16)=12 P(EE1)=14 ( ; out of four ‘1’ option is correct) P(EE2)=18 P(EE3)=1 (Becuase he knows the answer) P(E3E)=P(EE3):P(E3)E3P(EE9).PLEC)=1.1214.13+18.16+12.1 = 2429

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 make a guess is 13 and the probability that he copies the answer is 16. The probability that his answer is correct given that he copied it is 18 then the probability that he know the answer to the questions given that he correctly answered it is

E1 : gauss the answer. E2 : Copies the answer E3 : Known the answer, E : answer is correctP(E1)=13,P(E2)=16,P(E3)=1−(13+16)=12 P(EE1)=14 ( ; out of four ‘1’ option is correct) P(EE2)=18 P(EE3)=1 (Becuase he knows the answer) P(E3E)=P(EE3):P(E3)E3P(EE9).PLEC)=1.1214.13+18.16+12.1 = 2429

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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 make a guess is $\frac{1}{3}$ and the probability that he copies the answer is $\frac{1}{6}$ . The probability that his answer is correct given that he copied it is $\frac{1}{8}$ then the probability that he know the answer to the questions given that he correctly answered it is

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$\frac{25}{29}$

$\frac{24}{29}$

$\frac{27}{29}$

$\frac{28}{29}$

answer is C.

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Detailed Solution

$E_{1} : $ gauss the answer. $E_{2} : $ Copies the answer
$E_{3} : $ Known the answer, E : answer is correct

$P (E_{1}) = \frac{1}{3}, P (E_{2}) = \frac{1}{6}, P (E_{3}) = 1 - (\frac{1}{3} + \frac{1}{6}) = \frac{1}{2} P (\frac{E}{E_{1}}) = \frac{1}{4} (; o u t o f f o u r ‘ 1 ’ o p t i o n i s c o r r e c t) P (\frac{E}{E_{2}}) = \frac{1}{8} P (\frac{E}{E_{3}}) = 1 (B e c u a s e h e k n o w s t h e a n s w e r) \begin{array}{l} P (\frac{E_{3}}{E}) = \frac{P (\frac{E}{E_{3}}) : P (E_{3})}{E^{3} P (\frac{E}{E_{9}}) . P L E C)} = \frac{1. \frac{1}{2}}{\frac{1}{4} . \frac{1}{3} + \frac{1}{8} . \frac{1}{6} + \frac{1}{2} .1} \\ = \frac{24}{29} \end{array}$