# In a test an examinee either guesses or copies or knows the answer to a multiple choice question with four choices.

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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 makes 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 $\frac{1}{8}$ The probability that his answer is correct, given that he guessed it is $\frac{1}{4}$ The probability that they knew the answer to the questions given that he correctly answered it is

(a) $\frac{24}{31}$

(b) $\frac{31}{24}$

(c) $\frac{24}{29}$

(d) $\frac{29}{24}$

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Answer : (c) $\frac{24}{29}$

Let,

E : Event examinee answers correctly

Given, P(E1) = $\frac{1}{3}$ , P(E2) = $\frac{1}{6}$

∴  P(E3) = 1 – (P(E1) + P(E3)) = 1- $\big(\frac{1}{3}+\frac{1}{6}\big)$ = $\frac{1}{2}$

Given, $P(E/E_1) =\frac{1}{4},P(E/E_2) =\frac{1}{8},P(E/E_3)=1$

∴ Required probability =  P(E3/E)

$\frac{P(E_3).P(E/E_3)}{P(E_1).P(E/E_1)+P(E_2).P(E/E_2)+P(E_3).P(E/E_3)}$

$= \frac{\frac{1}{2}.1}{\frac{1}{3}\times \frac{1}{4}+\frac{1}{6}\times \frac{1}{8}+\frac{1}{2}\times 1}$   = $\frac{\frac{1}{2}}{\frac{1}{12}+\frac{1}{48}+\frac{1}{2}}$ = $\frac{\frac{1}{2}}{\frac{4+1+24}{48}}$ = $\frac{\frac{1}{2}}{\frac{29}{48}}$  $=\frac{24}{29}$

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