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 a student knows the answer given that the answered it correctly?

Answer 1

Let $E_1$ : the event that the student knows the answer and $E_2$ the event that the student guesses the answer.

Therefore $E_1$ and $E_2$ are mutually exclusive and exhaustive.

$\therefore P\left(E_1\right)=\frac{3}{4}andP\left(E_2\right)=\frac{1}{4}$

Let E: the answer is correct.

The probability that the student answered correctly, given that he knows the answer, is 1 i.e., $P\left(\frac{E}{E_1}\right)=1$

Probability that the students answered correctly, given that the he guessed, is $\frac{1}{4}i.e.,P\left(\frac{E}{E_2}\right)=\frac{1}{4}$.

By using Baye' s theorem, we obtain

$P\left(\frac{E_1}{E}\right)=\frac{P\left(\frac{E}{E_1}\right)P\left(E_1\right)}{P\left(\frac{E}{E_1}\right)P\left(E_1\right)+P\left(\frac{E}{E_2}\right)P\left(E_2\right)}$

=$\frac{1\times \frac{3}{4}}{1\times \frac{3}{4}+\frac{1}{4}\times \frac{1}{4}}=\frac{\frac{3}{4}}{\frac{3}{4}+\frac{1}{16}}=\frac{\frac{3}{4}}{\frac{12+1}{16}}=\frac{3}{4}\times \frac{16}{13}=\frac{12}{13}$

Note: If two events are not mutually exclusive/exhaustive, then you do not use Baye's theorem.