Question

In a test an examinee either guesses or copies or knows the answer to a multiple choice question with four choices. The probability that he make, a guess is $\frac{1}{3}$ and the probability that he copies the answer is $\frac{1}{6}$. The probability that his answer is correct given that he copied it is $\frac{1}{8}$. Find the probability that he knew the answer to the question, given that he correctly answered it.

Answer 1

P(g) = probability of guessing $=\frac{1}{3}$

P(c) = probability of copying $=\frac{1}{6}$

P(k) = probability of knowing $=1-\frac{1}{3}-\frac{1}{6}=\frac{1}{2}$

Probability of answering correctly during guess $=1/4$ (one out of four choices)

Probability of answering correctly during copy $=1/8$

Probability of answering correctly when he knew answer to the question (by solving) = 1 (As he knows answer he will answer it correctly).

Hence using Bayes formula (or theorem)

The probability that he knew the answer to the question, given that he answered it correctly

$=\frac{(1\times \frac{1}{2})}{(\frac{1}{3}\times \frac{1}{4}+\frac{1}{6}\times \frac{1}{8}+\frac{1}{2}\times 1)}$

$=\frac{\frac{1}{2}}{(\frac{1}{12}+\frac{1}{48}+\frac{1}{2})}=\frac{24}{29}$