Question

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 probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability $\dfrac{1}{4}$ . What is the probability that a student knows the answer given that he answered it correctly?

Answer 1

Hint: In order to solve the given problem; use the Bayes theorem of probability. First find the probability for different conditions keeping in mind the practical conditions and finally substitute all the values in the formula for Bayes theorem to find the probability.

Complete step by step answer:
Let ${E_a}$ be the event that the student knows the answer.
And let ${E_b}$ be the event that the student guesses the answer.
We know the values of these probabilities are:
$
   \Rightarrow P\left( {{E_a}} \right) = \dfrac{3}{4} \\
   \Rightarrow P\left( {{E_b}} \right) = 1 - P\left( {{E_a}} \right) = 1 - \dfrac{3}{4} \\
   \Rightarrow P\left( {{E_b}} \right) = \dfrac{1}{4} \\
 $
And let $M$ be the event that the answer given by a student is correct.
Practically we know that the student will answer the question correctly if he knows the answer. So, this can be represented as:
$ \Rightarrow P\left( {M|{E_a}} \right) = 1$
According to the question, we have the probability of the question is correct if the student guesses the answer is $\dfrac{1}{4}$
This can be represented as:
$ \Rightarrow P\left( {M|{E_b}} \right) = \dfrac{1}{4}$
We have to find the probability that the student knows the answer when we have the answer is correct.
This can be mathematically represented as:
$P\left( {{E_a}|M} \right)$
Let us use the Bayes theorem to find the probability.
The formula for Bayes theorem for two conditions is
$ \Rightarrow P\left( {{E_a}|M} \right) = \dfrac{{P\left( {{E_a}} \right) \cdot P\left( {M|{E_a}} \right)}}{{P\left( {{E_a}} \right) \cdot P\left( {M|{E_a}} \right) + P\left( {{E_b}} \right) \cdot P\left( {M|{E_b}} \right)}}$
Let us substitute the values in the Bayes theorem to find the result.
$ \Rightarrow P\left( {{E_a}|M} \right) = \dfrac{{\dfrac{3}{4} \cdot 1}}{{\left( {\dfrac{3}{4} \cdot 1} \right) + \left( {\dfrac{1}{4} \cdot \dfrac{1}{4}} \right)}}$
Let us solve the given term to find the probability.
$
   \Rightarrow P\left( {{E_a}|M} \right) = \dfrac{{\dfrac{3}{4}}}{{\dfrac{3}{4} + \dfrac{1}{{16}}}} \\
   \Rightarrow P\left( {{E_a}|M} \right) = \dfrac{{\dfrac{3}{4}}}{{\dfrac{{13}}{{16}}}} \\
   \Rightarrow P\left( {{E_a}|M} \right) = \dfrac{3}{4} \times \dfrac{{16}}{{13}} \\
   \Rightarrow P\left( {{E_a}|M} \right) = \dfrac{{12}}{{13}} \\
 $
Hence, the probability that a student knows the answer given that he answered it correctly is $\dfrac{{12}}{{13}}$

Note: In order to solve such problems with different cases and conditional probability, students must try to use different formulas like Bayes theorem to make the problem easier. The problem can also be solved by simple methods but that will be even more difficult to solve. Bayes theorem is a formula that explains how, when given proof, to change the probability of the hypotheses. It actually stems from the axioms of conditional probability, which can be used to effectively argue for a wide variety of concerns including changes to beliefs.