Question

In answering a question on a multiple choice test a student either knows the answeror guesses.Let \(\frac{3}{4}\) be the probability that he knows the answer and \(\frac{1}{4}\) be the probability that he guesses.Assuming that the student who guessws the answer,will be correct with probability \(\frac{1}{4}\).What is the probability that a student knows the answer given that he answered it correctly?

Answer 1

The correct answer is: \(\frac{12}{13}\)
Let \(E_1\)=the examiee knows the answer, \(E_2\)=the examinee guesses the answer and A=the examinee answers correctly
Now \(P(E_1)=\frac{3}{4},P(E_2)=\frac{1}{4},\)
Since \(E_1\) and \(E_2\) are mutually exclusive events and exhaustive events, and if \(E_2\) has already occurred, then the examinee guesses,
Therefore the probability that he answers correctly given that he has made a guess is \(\frac{1}{4}\) i.e., \(P(A|E_2)=\frac{1}{4}\)
And \(P(A|E_2)\)=P(answers correctly given that he knew the answer)=1
Therefore,by Bayes' theorem,
\(P(E_1|A)=\frac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1)+P(E_2)P(A|E_2)}\)
\(=\frac{\frac{3}{4}.1}{\frac{3}{4}.1+\frac{1}{4}.\frac{1}{4}}\)
\(=\frac{\frac{3}{4}}{\frac{3}{4}+\frac{1}{16}}\)
\(=\frac{\frac{3}{4}}{\frac{13}{16}}\)
\(=\frac{12}{13}\)