In answering a question on a multiple choice test, a student either knows the answer or guesses. Let $\cfrac{3}{4}$ be the probability that he knows the answer and $\cfrac{1}{4}$ the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability $\cfrac{1}{4}$ . What is the probability that the student knows the answer given that he answered it correctly?

.: Let ${E_1}$ : `the student knows the answer'

And ${E_2}$ : `the student guesses the answer'
,
$\Rightarrow$ $P\left( {{E_1}} \right) = \cfrac{3}{4}$

and $P\left( {{E_2}} \right) = \cfrac{1}{4}$
Let $A$ : 'the answer is correct',

then $P\left( {A|{E_1}} \right) = 1$ and

$P\left( {A|{E_2}} \right) = \cfrac{1}{4}$

( when the student knows the answer,

it is a sure event that the answer is correct)
Hence, the required probability is

$P\left( {{E_1}|A} \right) = \cfrac{{P\left( {A|{E_1}} \right)P\left( {{E_1}} \right)}}{{P\left( {A|{E_1}} \right)P\left( {{E_1}} \right) + P\left( {A|{E_2}} \right)P\left( {{E_2}} \right)}}$

$= \cfrac{{1 \times \cfrac{3}{4}}}{{1 \times \cfrac{3}{4} + \cfrac{1}{4} \times \cfrac{1}{4}}} = \cfrac{{\cfrac{3}{4}}}{{\cfrac{3}{4} + \cfrac{1}{{16}}}}$

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