An urn contains five balls. Two balls are drawn and found to be white. The probability that all the balls are white is :

Solution :
Let $${A_i}\left( {i = 2,\,3,\,4,\,5} \right)$$     be the event that urn contains $$2,\,3,\,4,\,5$$    white balls and let $$B$$ be the event that two white balls have been drawn then we have to find $$P\left( {\frac{{{A_5}}}{B}} \right).$$
Since the four events $${A_2},\,{A_3},\,{A_4}$$   and $${A_5}$$ are equally likely we have $$P\left( {{A_2}} \right) = P\left( {{A_3}} \right) = P\left( {{A_4}} \right) = P\left( {{A_5}} \right) = \frac{1}{4}.$$
$$P\left( {\frac{B}{{{A_2}}}} \right)$$  is probability of event that the urn contains $$2$$ white balls and both have been drawn.
$$\eqalign{ & \therefore \,P\left( {\frac{B}{{{A_2}}}} \right) = \frac{{{}^2{C_2}}}{{{}^5{C_2}}} = \frac{1}{{10}} \cr & {\text{Similarly,}} \cr & P\left( {\frac{B}{{{A_3}}}} \right) = \frac{{{}^3{C_2}}}{{{}^5{C_2}}} = \frac{3}{{10}}, \cr & P\left( {\frac{B}{{{A_4}}}} \right) = \frac{{{}^4{C_2}}}{{{}^5{C_2}}} = \frac{3}{5}, \cr & P\left( {\frac{B}{{{A_5}}}} \right) = \frac{{{}^5{C_2}}}{{{}^5{C_2}}} = 1 \cr & {\text{By Bayes theorem,}} \cr & P\left( {\frac{{{A_5}}}{B}} \right) = \frac{{P\left( {{A_5}} \right)P\left( {\frac{B}{{{A_5}}}} \right)}}{{\left( {P\left( {{A_2}} \right)P\left( {\frac{B}{{{A_2}}}} \right) + P\left( {{A_3}} \right)P\left( {\frac{B}{{{A_3}}}} \right) + P\left( {{A_4}} \right)P\left( {\frac{B}{{{A_4}}}} \right) + P\left( {{A_5}} \right)P\left( {\frac{B}{{{A_5}}}} \right)} \right)}} \cr & = \frac{{\frac{1}{4}.1}}{{\frac{1}{4}\left[ {\frac{1}{{10}} + \frac{3}{{10}} + \frac{3}{5} + 1} \right]}} \cr & = \frac{{10}}{{20}} \cr & = \frac{1}{2} \cr} $$