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

Let $E_1$ and $E_2$ be the respective events that the student knows the answer and he guesses the answer.
Let $A$ be the event that the answer is correct.

$\therefore P\left(E_1\right)=\frac{3}{4}$

$P\left(E_2\right)=\frac{1}{4}$

The probability that the student answered correctly, given that he knows the answer, is 1.
$\therefore P\left(A|E_1\right)=1$
Probability that the student answered correctly, given that he guessed, is $\frac{1}{4}$
$\therefore P\left(A|E_2\right)=\frac{1}{4}$
The probability that the student knows the answer, given that he answered it correctly, is given by $P\left(E_1|A\right)$

By using Baye's theorem, we obtain
$P\left(E_1|A\right)=\frac{P\left(E_1\right)\cdot P\left(A|E_1\right)}{P\left(E_1\right)\cdot P\left(A|E_1\right)+P\left(E_2\right)\cdot P\left(A|E_2\right)}$

$=\frac{\frac{3}{4}\cdot 1}{\frac{3}{4}\cdot 1+\frac{1}{4}\cdot \frac{1}{4}}$

$=\frac{\frac{3}{4}}{\frac{3}{4}+\frac{1}{16}}$

$=\frac{\frac{3}{4}}{\frac{13}{16}}$

$=\frac{12}{13}=0.92$.