Answer:
Let\[{{E}_{1}}\] : the event that the student knows the answer and\[{{E}_{2}}\]: the event that the student guesses the answer. Therefore \[{{E}_{1}}\] and \[{{E}_{2}}\] are mutually exclusive and exhaustive. \[\therefore \] \[P({{E}_{1}})=\frac{3}{4}\,\,\text{and}\,\,P({{E}_{2}})=\frac{1}{4}\] Let E : the answer is correct. The probability that the student answered correctly, given that he knows the answer, is 1, i.e. \[P\left( \frac{E}{{{E}_{1}}} \right)=1\] Probability that the students answered correctly, given that the he guessed, is \[\frac{1}{4},\] i.e. \[P\left( \frac{E}{{{E}_{2}}} \right)=\frac{1}{4}.\] By using Baye's theorem, \[P\left( \frac{{{E}_{1}}}{E} \right)=\frac{P\left( \frac{E}{{{E}_{1}}} \right)P({{E}_{1}})}{P\left( \frac{E}{{{E}_{1}}} \right)P({{E}_{1}})+P\left( \frac{E}{{{E}_{2}}} \right)P({{E}_{2}})}\] \[=\frac{1\times \frac{3}{4}}{1\times \frac{3}{4}+\frac{1}{4}\times \frac{1}{4}}=\frac{\frac{3}{4}}{\frac{3}{4}+\frac{1}{16}}=\frac{\frac{3}{4}}{\frac{12+1}{16}}\] \[=\frac{3}{4}\times \frac{16}{13}=\frac{12}{13}\]
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