Answer : (c) \(\frac{24}{29}\)
Let,
E1: Examinee guesses the answer
E2: Examinee copies the answer
E3: Examinee knows the answer
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)
= \(\frac{P(E_3).P(E/E_3)}{P(E_1).P(E/E_1)+P(E_2).P(E/E_2)+P(E_3).P(E/E_3)} \)
\(= \frac{\frac{1}{2}.1}{\frac{1}{3}\times \frac{1}{4}+\frac{1}{6}\times \frac{1}{8}+\frac{1}{2}\times 1}\) = \(\frac{\frac{1}{2}}{\frac{1}{12}+\frac{1}{48}+\frac{1}{2}}\) = \(\frac{\frac{1}{2}}{\frac{4+1+24}{48}}\) = \(\frac{\frac{1}{2}}{\frac{29}{48}}\) \(=\frac{24}{29}\)