 # In answering a question on a multiple choice test, a student either knows the answer or guesses.

Question:

In answering a question on a multiple choice test, a student either knows the answer or guesses. Let $\frac{3}{4}$ be the probability that he knows the answer and $\frac{1}{4}$ be the probability

that he guesses. Assuming that a student who guesses at the answer will be correct with probability What is the probability that the student knows the answer given that he answered it correctly?

Solution:

Let E1 and E2 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.

$\therefore P\left(A \mid E_{1}\right)=1$

Probability that the student answered correctly, given that he guessed, is $\frac{1}{4}$.

$\therefore \mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{2}\right)=\frac{1}{4}$

The probability that the student knows the answer, given that he answered it correctly, is given by $\mathrm{P}\left(\mathrm{E}_{\mid} \mid \mathrm{A}\right)$.

By using Bayes’ theorem, we obtain

$P\left(E_{1} \mid A\right)=\frac{P\left(E_{1}\right) \cdot P\left(A \mid E_{1}\right)}{P\left(E_{1}\right) \cdot P\left(A \mid E_{1}\right)+P\left(E_{2}\right) \cdot P\left(A \mid E_{2}\right)}$

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

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

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

$=\frac{12}{13}$