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 $\frac{1}{4}$. What is the probability that the student knows the answer given that he answered it correctly?

Answer

Let $E_1$ : the student knows the answer, then $, P(E_1) = \frac{3}{4}$
$E_2$ : the student guesses the answer, then $, P(E_2) = \frac{1}{4}$
Let $A$: the answer is correct.
$P(\frac {A}{{E_1}}) = 1, P(\frac {A}{{E_2}}) = \frac{1}{4}$
Hence, by Baye's theorm, we have,
$P(\frac {E_1}{A}) = \frac{{P({E_1})P(A/{E_1})}}{{P({E_1})P(A/{E_1}) + P({E_2})P(A/{E_2})}}$
$ = \frac{{1 \times \frac{3}{4}}}{{1 \times \frac{3}{4} + \frac{1}{4} \times \frac{1}{4}}} = \frac{{12}}{{13}}$

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