Let define the following events:\\\\({ E }_{ 1 }:\\\\) Examine guesses the answer to the question.\\\\({ E }_{ 2 }:\\\\) Examine copies the answer to the question.\\\\({ E }_{ 3 }:\\\\) Examine know the answer to the question.\\\\(E:\\\\) answer is correctHere \\\\(\\displaystyle P\\left( { E }_{ 1 } \\right) =\\frac { 1 }{ 3 } ,P\\left( { E }_{ 2 } \\right) =\\frac { 1 }{ 6 } \\\\)\\\\(\\displaystyle \ herefore P\\left( { E }_{ 3 } \\right) =1-\\left( \\frac { 1 }{ 3 } +\\frac { 1 }{ 6 } \\right) =\\frac { 1 }{ 2 } \\\\)Since the question is a multiple choice question with four choice so the probability that the answer is correct when is guessed is \\\\(\\displaystyle P\\left( \\frac { E }{ { E }_{ 1 } } \\right) =\\frac { 1 }{ 4 } \\\\)Also the probability that his answer is correct, given that he copied it is \\\\(\\displaystyle \\frac { 1 }{ 8 } \\\\)i.e \\\\(\\displaystyle P\\left( \\frac { E }{ { E }_{ 2 } } \\right) =\\frac { 1 }{ 8 } \\\\)Moreover his answer is correct, given he knew the answer is a sure event, so \\\\(\\displaystyle P\\left( \\frac { E }{ { E }_{ 3 } } \\right) =1\\\\)Hence by Baye's theorem the required probability is\\\\(\\displaystyle P\\left( \\dfrac { { E }_{ 3 } }{ E } \\right) =\\dfrac { P\\left( \\dfrac { E }{ { E }_{ 3 } } \\right) .P\\left( { E }_{ 3 } \\right) }{ \\sum _{ i=1 }^{ 3 }{ P\\left( \\dfrac { E }{ { E }_{ i } } \\right) .P\\left( { E }_{ i } \\right) } } =\\dfrac { 1.\\dfrac { 1 }{ 2 } }{ \\dfrac { 1 }{ 4 } .\\dfrac { 1 }{ 3 } +\\dfrac { 1 }{ 8 } .\\dfrac { 1 }{ 6 } +\\dfrac { 1 }{ 2 } .1 } =\\dfrac { 24 }{ 29 } \\\\)