Question
In answering a question on a multiple choice questions test with four choices in each question, out of which only one is correct, a student either guesses or copies or knows the answer. The probability that he makes a guess $\frac{1}{4}$
and the probability the he copies is also $\frac{1}{4}$ The probability that the answer is correct, given that he copied it is $\frac{3}{4}$ Find the probability that he knows the answer to the question, given that he correctly answered it.

Answer

Let $E_1=$ Student guesses the answer
$E_2=$ Student copies the answer
$E_3 =$ Student knows the answer
$A =$ Student answers the question correctly.
$ P \left( E _1\right)=\frac{1}{4},$
$P \left( E _2\right)=\frac{1}{4},$
$P \left( E _3\right)=1-\left(\frac{1}{4}+\frac{1}{4}\right)=\frac{1}{2}$
$P \left( A \mid E _1\right)=\frac{1}{4}, P \left( A \mid E_2\right)=\frac{3}{4}, P \left( A \mid E_3\right)=1$
The required probability
$=P\left(E_3 \mid A\right)=\frac{P\left(E_3\right) \times P\left(A \mid E_3\right)}{\sum_{i=1}^3 P\left(E_i\right) \times P\left(A \mid E_i\right)}$
$=\frac{\frac{1}{2} \times 1}{\frac{1}{4} \times \frac{1}{4}+\frac{1}{4} \times \frac{3}{4}+\frac{1}{2} \times 1}$
$=\frac{1}{\frac{1}{8}+\frac{3}{8}+1}=\frac{8}{12}=\frac{2}{3}$

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