Bag I contains 3 red and 4 black balls and bag II contains 4 red and 4 black balls. Two balls are transferred at random from bag I to bag II and then a ball is drawn from bag II. The ball so drawn is found to be red colour, then what is the probability that the transferred balls were both black.

Event $E_1=\left\{\begin{array}{l}drawntwoballsfrom\\ bagIisbothareblack\end{array}\right\}$

$P\left(E_1\right)=\frac{{}^{4}C_2}{{}^{7}C_2}=\frac{2}{7}$

Event $E_2=\left\{botharered\right\}\Rightarrow P\left(E_2\right)=\frac{{}^{3}C_2}{{}^{7}C_2}=\frac{1}{7}$

Event $E_3=\{oneblack,onered\}\Rightarrow P\left(E_3\right)=\frac{{}^{3}C_1\times {}^{4}C_1}{{}^{7}C_2}=\frac{4}{7}$

Let the event $=\left\{\text{After transforming balls from I to II the drawn ball from II is red}\right\}$

$P(A/E_1)=\frac{{}^{4}C_1}{{}^{10}C_1}=\frac{2}{5},P(A/E_2)=\frac{{}^{6}C_1}{{}^{10}C_1}=\frac{3}{5}$

$P(A/E_3)=\frac{5C_1}{10C_1}=\frac{1}{2}$

$\begin{array}{l}\therefore P(E_1/A)=\frac{P\left(E_1\right)P(A/E_1)}{P\left(E_1\right)P(A/E_1)+P\left(E_2\right)P(A/E_2)+P\left(E_3\right)P(A/E_3)}\\ =\frac{\frac{2}{7}\left(\frac{2}{5}\right)}{\frac{2}{7}\left(\frac{2}{5}\right)+\frac{1}{7}\left(\frac{3}{5}\right)+\frac{4}{7}\left(\frac{1}{2}\right)}\\ =\frac{4}{4+3+10}=\frac{4}{17}\end{array}$