Suppose we have four boxes $A,B,C$ and $D$ containlng oloured marbles as given below:
Box Marble colour
Red White Black
$A$ 1 6 3
$B$ 6 2 2
$C$ 8 1 1
$D$ 0 6 4
One of the boxes has been selected at random and a single marble is drawn from it If the marble is red, what is the probability that it was drawn from box $A$ ?, box $B$ ?, box $C$ ?

.: Let ${E_1}:$ box $A$ be selected
${E_2}:$ box ${\rm{B}}$ be selected; ${E_3}:$

box ${\rm{C}}$ be selected; ${E_4}$ box ${\rm{D}}$ be selected, ${\rm{A}}$

: Red ball be
-drawn,

$P\left( {{E_1}} \right) = \cfrac{1}{4},P\left( {{E_2}} \right) = \cfrac{1}{4},P\left( {{E_3}} \right) = \cfrac{1}{4},P\left( {{E_4}} \right) = \cfrac{1}{4}$

$P\left( {A|{E_1}} \right) = \cfrac{1}{{10}},P\left( {A|{E_2}} \right) = \cfrac{6}{{10}},P\left( {A|{E_3}} \right) = \cfrac{8}{{10}},P\left( {A|{E_4}} \right) = \cfrac{0}{{10}}$

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

$+ P\left( {{E_3}} \right)P\left( {A|{E_3}} \right) + P\left( {{E_4}} \right)P\left( {A|{E_4}} \right)$

$= \cfrac{{\cfrac{1}{4} \times \cfrac{1}{{10}}}}{{\cfrac{1}{4} \times \cfrac{1}{{10}} + \cfrac{1}{4} \times \cfrac{6}{{10}} + \cfrac{1}{4} \times \cfrac{8}{{10}} + \cfrac{1}{4} \times \cfrac{0}{{10}}}} = \cfrac{1}{{(1 + 6 + 8 + 0)}} = \cfrac{1}{{15}}$

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

$+ P\left( {{E_3}} \right)P\left( {A|{E_3}} \right) + P\left( {{E_4}} \right)P\left( {A|{E_4}} \right)$

$= \cfrac{{\cfrac{1}{4} \times \cfrac{6}{{10}}}}{{\cfrac{1}{4} \times \cfrac{1}{{10}} + \cfrac{1}{4} \times \cfrac{6}{{10}} + \cfrac{1}{4} \times \cfrac{8}{{10}} + \cfrac{1}{4} \times \cfrac{0}{{10}}}} = \cfrac{6}{{1 + 6 + 8 + 0}} = \cfrac{6}{{15}} = \cfrac{2}{5}$

$P\left( {{E_3}|A} \right) = \cfrac{{P\left( {{E_3}} \right)P\left( {A|{E_3}} \right)}}{{P\left( {{E_1}} \right)P\left( {A|{E_1}} \right) + P\left( {{E_2}} \right)P\left( {A|{E_2}} \right)}}$

$+ P\left( {{E_3}} \right)P\left( {A|{E_3}} \right) + P\left( {{E_4}} \right)P\left( {A|{E_4}} \right)$

$= \cfrac{{\cfrac{1}{4} \times \cfrac{8}{{10}}}}{{\cfrac{1}{4} \times \cfrac{1}{{10}} + \cfrac{1}{4} \times \cfrac{6}{{10}} + \cfrac{1}{4} \times \cfrac{8}{{10}} + \cfrac{1}{4} \times \cfrac{0}{{10}}}} = \cfrac{8}{{1 + 6 + 8 + 0}} = \cfrac{8}{{15}}$

So, $P\left( {{E_1}|A} \right) = \cfrac{1}{{15}},P\left( {{E_2}|A} \right) = \cfrac{2}{5},P\left( {{E_3}|A} \right) = \cfrac{8}{{15}}$