Bayes Theorem describes the probability of an event A, based on prior knowledge that event B has already happened. 

Suppose there are two bags \({{\rm{B}}_{\rm{1}}}\) and \({{\rm{B}}_{\rm{2}}}\). \({{\rm{B}}_{\rm{1}}}\) contains 2 red and 8 black balls, while the bag \({{\rm{B}}_{\rm{2}}}\) contains 7 red and 3 black balls. We selected a bag randomly and a ball in taken from it and this ball is found to be red, now we want to calculate the probability that the ball is taken from the bag \({{\rm{B}}_{\rm{1}}}\).  Intuitively, the chances that the ball came from the bag \({{\rm{B}}_{\rm{1}}}\) is less as this bag has only 2 red balls. Here the knowledge that the ball is red is affecting the chances of a bag being selected.

The red ball coming from the Bag \({{\rm{B}}_{\rm{1}}}\)= \(\cfrac{1}{2} \times \cfrac{2}{{10}}\)

The red ball coming from the Bag \({{\rm{B}}_{\rm{2}}}\) = \(\cfrac{1}{2} \times \cfrac{7}{{10}}\) 

\(P({\rm{Bag}}\;{{\rm{B}}_{\rm{1}}}{\rm{|Red}}\;{\rm{ball}}\;{\rm{is}}\;{\rm{selected}}) = \cfrac{{P({\rm{Red}}\;{\rm{ball}}\;{\rm{from}}\;{{\rm{B}}_{\rm{1}}})}}{{P({\rm{red}}\;{\rm{ball)}}}}\)

\( = \cfrac{{\cfrac{1}{2} \times \cfrac{2}{{10}}}}{{\cfrac{1}{2} \times \cfrac{2}{{10}} + \cfrac{1}{2} \times \cfrac{7}{{10}}}} = \cfrac{2}{9}\)

Let us take an example:

Solution: There are five bags divided into two groups, if \({q_1}\) and \({q_2}\) are the probabilities that first or second group is selected. Here \({q_1} = \frac{3}{5},\;{q_2} = \frac{2}{5}\)

A bag is selected from 1st group and the chances of drawing a black ball is \({p_1} = \frac{2}{7}\), 

If it is selected from the second group, then probability is \({p_2} = \frac{4}{5}\). 

Required probability P(E) \( = \cfrac{{\frac{3}{5} \times \frac{2}{7}}}{{\frac{3}{5} \times \frac{2}{7} + \frac{2}{5} \times \frac{4}{5}}} = \cfrac{{\frac{6}{{35}}}}{{\frac{6}{{35}} + \frac{8}{{25}}}} = \cfrac{{15}}{{43}}\)

Example 01:  A man is known to speak the truth 3 out of 4 times. He throws a dice and reports that it is a six. Find the probability that it is actually six. 

Example 02:  A student is taking a test having questions in 4 choices MCQ format and he is not well prepared, the probability that he knows the answer of a random question is only 12%. In case he does not know the answer, he makes a random guess. If he Answer a randomly selected question correctly find the probability that he knew the answer of the question already.