Bag $I$ contain 3 red and 4 black balls and Bag $II$ contains 4 red and 5 black balls. One ball is transferred from Bag $I$ to Bag $II$ and then a ball is drawn from Bag $II$ . The ball so drawn is found to be red in colour. Find the probability that the transferred ball is black.Choose the correct answer in each of the following

.: ${E_1}:$ Red ball is transferred from Bag $I$ to Bag $II$

${E_2}:$ Black ball is transferred from Bag $I$ to Bag $II$

and $A:$ Red ball is drawn from Bag $II$

$\therefore$ $P\left( {{E_1}} \right) = \cfrac{3}{7},P\left( {{E_2}} \right) = \cfrac{4}{7}$

Also, $P\left( {A|{E_1}} \right) = \cfrac{5}{{10}} = \cfrac{1}{2}$ and $P\left( {A|{E_2}} \right) = \cfrac{4}{{10}} = \cfrac{2}{5}$

By Baye's theorem,
$P\left( {{E_2}|A} \right) = \cfrac{{P\left( {{E_2}} \right)P\left( {A|{E_2}} \right)}}{{P\left( {{E_1}} \right)P\left( {A|{E_1}} \right) + P\left( {{E_2}} \right)P\left( {A|{E_2}} \right)}}$

$= \cfrac{{\left( {\cfrac{4}{7}} \right)\left( {\cfrac{4}{{10}}} \right)}}{{\left( {\cfrac{3}{7}} \right)\left( {\cfrac{1}{2}} \right) + \left( {\cfrac{4}{7}} \right)\left( {\cfrac{2}{5}} \right)}} = \cfrac{{16}}{{70}} \times \cfrac{{70}}{{31}} = \cfrac{{16}}{{31}}$