**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(E_{1}) = \(\frac{1}{3}\) , P(E_{2}) = \(\frac{1}{6}\)

∴ P(E_{3}) = 1 – (P(E_{1}) + P(E_{3})) = 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(E_{3}/E)

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

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