Question

In answering a question on a MCQ test with 4 choices per question, a student knows the answer guesse or copies the answer.Let 1/2 be the probability that he knows the answer,1/4 be the probability that he guesses & 1/4 that he copies it. Assuming that a student who copies the answer will be correct with the probability 3/4 what is the probability that the student knows the answer given that he answered it correctly

Answer 1

Answer:

$\boxed{\sf \: Required\:probability\:is\:\dfrac{2}{3} \: }$

Step-by-step explanation:

Let assume the following events

E₁ : Student guess the answer to a question.

E₂ : Student copies the answer to a question.

E₃ : Student knows the amswer to a question.

E : He answered the question correctly.

Now,

$\sf \: P(E_1) = \dfrac{1}{4}$

$\sf \: P(E_2) =  \dfrac{1}{4}$

$\implies\sf \: P(E_3) = \dfrac{1}{2}$

Now, Further given that

$\sf \: P(E  \: \mid \: E_1) = \dfrac{1}{4}$

$\sf \: P(E  \: \mid \: E_2) = \dfrac{3}{4}$

$\sf \: P(E  \: \mid \: E_3) = 1$

Now, By definition of Bayes Theorem, we have

$\sf P(E_3\mid \: E) = \dfrac{P(E_3) . P(E  \mid E_3)}{P(E_1) . P(E  \mid E_1) + P(E_2) . P(E  \mid E_2)+ P(E_3) . P(E  \mid E_3)}$

So, on substituting the values, we get

$\sf P(E_3 \mid \: E) = \dfrac{\dfrac{1}{2}  \times 1}{\dfrac{1}{4}  \times \dfrac{1}{4}  + \dfrac{1}{4}  \times \dfrac{3}{4} + \dfrac{1}{2} \times 1}$

$\sf P(E_3 \mid \: E) = \dfrac{8}{1 + 3 + 8}$

$\sf\implies \sf P(E_3 \mid \: E) = \dfrac{2}{3}$

Hence,

$\sf\implies \sf Required\:probability\:is\:\dfrac{2}{3}$