Question

In a test an examinee either guesses or copies or knows the answer to a multiple choice question with 4 choices. The probability that he makes a guess is 1/3 & the probability that he copies the answer is 1/6. The probability that his answer is correct given that he copied it, is 1/8. Find the probability that he knew the answer to the question given that he correctly answered it.
If expressed in the form of $a/b$ (simplest form), $b-a=?$

Answer 1

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 correct
Here $P\left(E_1\right)=\frac{1}{3},P\left(E_2\right)=\frac{1}{6}$
$\therefore P\left(E_3\right)=1-(\frac{1}{3}+\frac{1}{6})=\frac{1}{2}$
Since the question is a multiple choice question with four choice so the probability that the answer is correct
when is guessed is $P\left(\frac{E}{E_1}\right)=\frac{1}{4}$
Also the probability that his answer is correct, given that he copied it is $\frac{1}{8}$
i.e $P\left(\frac{E}{E_2}\right)=\frac{1}{8}$
Moreover his answer is correct, given he knew the answer is a sure event, so $P\left(\frac{E}{E_3}\right)=1$
Hence by Baye's theorem the required probability is
$P\left(\frac{E_3}{E}\right)=\frac{P\left(\frac{E}{E_3}\right).P\left(E_3\right)}{{\sum }_{i=1}^{3}P\left(\frac{E}{E_i}\right).P\left(E_i\right)}=\frac{1.\frac{1}{2}}{\frac{1}{4}.\frac{1}{3}+\frac{1}{8}.\frac{1}{6}+\frac{1}{2}.1}=\frac{24}{29}$