Question

In answering a question on a multiple choice test, a student either knows the answer or guesses. Let 3/4 be the probability that he knows the answer and 1/4 be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability 1/4. What is the probability that the student knows the answer given that he answered it correctly?

Answer 1

Answer:

The probability that the student knows the answer given that he answered it correctly is $\dfrac{12}{13}$.

Step-by-step explanation:

Let assume the following events

E₁ : Student knows the answer to a question.

E₂ : Student guess the answer to a question.

E : He answered the question correctly.

Now,

$\sf \: P(E_1) = \dfrac{3}{4}$

$\sf \: P(E_2) =  \dfrac{1}{4}$

Now, Further given that

$\sf \: P(E  \: \mid \: E_1) = 1$

$\sf \: P(E  \: \mid \: E_2) = \dfrac{1}{4}$

Now, By definition of Bayes Theorem, we have

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

So, on substituting the values, we get

$\sf P(E_1 \mid \: E) = \dfrac{\dfrac{3}{4}  \times 1}{\dfrac{3}{4}  \times 1  + \dfrac{1}{4}  \times \dfrac{1}{4}}$

$\sf P(E_1 \mid \: E) = \dfrac{12}{12 + 1}$

$\sf\implies \sf P(E_1 \mid \: E) = \dfrac{12}{13}$

Thus,

$\sf\implies \sf Required\:probability\:is\:\dfrac{12}{13}$

Hence,

The probability that the student knows the answer given that he answered it correctly is $\dfrac{12}{13}$.