In answering a question on a multiple-choice test, a student either knows the answer or guesses. Let $\dfrac{3}{4}$ be the probability that he knows the answer and $\dfrac{1}{4}$ be the possibility that he guesses. Assuming that a student who guesses the answer will be correct with the probability $\dfrac{1}{4}$. What is the possibility that the student knows the answer given that he answered correctly?

Hint: We can solve the given question using Bayes theorem, $\text{P}\left( \dfrac{\text{A}}{\text{C}} \right)=\dfrac{\text{P}\left( \text{A} \right)\text{.P}\left( \dfrac{\text{C}}{\text{A}} \right)}{\text{P}\left( \text{A} \right).\text{P}\left( \dfrac{\text{C}}{\text{A}} \right)+\text{P}\left( \text{B} \right).\text{P}\left( \dfrac{\text{C}}{\text{B}} \right)}$ where,$\text{P}\left( \text{A} \right)$ is the probability that student knows the answer $\text{P}\left( \text{B} \right)$ is the probability that the student guesses the answer$\text{P}\left( \dfrac{\text{C}}{\text{A}} \right)$ is the probability that the student answered correctly if he knows the answer, $\text{P}\left( \dfrac{\text{C}}{\text{B}} \right)$ is the probability that the student answered correctly if he guesses and $\text{P}\left( \dfrac{\text{A}}{\text{C}} \right)$ is the probability that the student knows the answer given he answered correctly. Put the given values and simplify to get the answer.

Complete step-by-step answer:
 Let’s consider, A be the event that the student knows the answer and B be the event that the student guesses. Let C be the event that the answer is correct. Then according to the question,
Given, the probability that the student knows the answer$\text{P}\left( \text{A} \right)=\dfrac{3}{4}$.
The probability that the student guesses the answer $\text{P}\left( \text{B} \right)=\dfrac{1}{4}$ .
The probability that the student answered correctly given he knows the answer is $1$ as there is only one correct answer.
$\Rightarrow \text{P}\left( \dfrac{\text{C}}{\text{A}} \right)=1$
(Given) The probability that the student answered correctly, if he guessed is$\text{P}\left( \dfrac{\text{C}}{\text{B}} \right)=\dfrac{1}{4}$ . Then the probability that the student knows the answer, given that he answered correctly is given by$\text{P}\left( \dfrac{\text{A}}{\text{C}} \right)$
We know that by baye’s theorem, $\text{P}\left( \dfrac{\text{A}}{\text{C}} \right)=\dfrac{\text{P}\left( \text{A} \right)\text{.P}\left( \dfrac{\text{C}}{\text{A}} \right)}{\text{P}\left( \text{A} \right).\text{P}\left( \dfrac{\text{C}}{\text{A}} \right)+\text{P}\left( \text{B} \right).\text{P}\left( \dfrac{\text{C}}{\text{B}} \right)}$
On putting the given values, we get-
$\Rightarrow \text{P}\left( \dfrac{\text{A}}{\text{C}} \right)=\dfrac{\dfrac{3}{4}.1}{\dfrac{3}{4}.1+\dfrac{1}{4}.\dfrac{1}{4}}=\dfrac{\dfrac{3}{4}}{\dfrac{3}{4}+\dfrac{1}{16}}=\dfrac{\dfrac{3}{4}}{\dfrac{12+1}{16}}$
$\Rightarrow \text{P}\left( \dfrac{\text{A}}{\text{C}} \right)=\dfrac{\dfrac{3}{4}}{\dfrac{13}{16}}=\dfrac{3}{4}\times \dfrac{16}{13}=\dfrac{3\times 4}{13}=\dfrac{12}{13}$
Hence, the probability that the student knows the answer, given that he answered correctly is $\dfrac{12}{13}$.

Note: Bayes theorem is a formula that is used for determining conditional probability. It is a way of finding a probability when we know certain other probabilities. In the question, the events A and B are mutually exclusive and exhaustive meaning (A∩B) =∅ if Sample set S= (A∪B). The students may get confused that how we got-
 $\Rightarrow \text{P}\left( \dfrac{\text{C}}{\text{A}} \right)=1$ . Since there is only one correct answer and the student knows the answer so the probability that he answered correctly will also be $1$.