question_answer

In answering a question on a multiple choice test, a student either knows the answer or guesses. Let 3/4 be the probability that he knows the answer and 1/4 be the probability that he guesses. Assuming that, a student who guesses at the answer will be correct with probability 1/4. What is the probability that a student knows the answer given that he answered it correctly?

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, \[P\left( \frac{{{E}_{1}}}{E} \right)=\frac{P\left( \frac{E}{{{E}_{1}}} \right)P({{E}_{1}})}{P\left( \frac{E}{{{E}_{1}}} \right)P({{E}_{1}})+P\left( \frac{E}{{{E}_{2}}} \right)P({{E}_{2}})}\]             \[=\frac{1\times \frac{3}{4}}{1\times \frac{3}{4}+\frac{1}{4}\times \frac{1}{4}}=\frac{\frac{3}{4}}{\frac{3}{4}+\frac{1}{16}}=\frac{\frac{3}{4}}{\frac{12+1}{16}}\]             \[=\frac{3}{4}\times \frac{16}{13}=\frac{12}{13}\]