Bag I contains 3 red, 4 black and 3 white balls and Bag II contains 2 red, 5 black and 2 white 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 black in colour. Then the probability, that the transferred ball is red, is :

Let $E \rightarrow$ Ball drawn from Bag II is black. <br/><br/> $E_{R} \rightarrow$ Bag I to Bag II red ball transferred. <br/><br/> $E_{B} \rightarrow$ Bag I to Bag II black ball transferred. <br/><br/> $E_{w} \rightarrow$ Bag I to Bag II white ball transferred. <br/><br/> $P\left(E_{R} / E\right)=\frac{P\left(E / E_{R}\right) \cdot P\left(E_{R}\right)}{P\left(E / E_{R}\right) P\left(E_{R}\right)+P\left(E / E_{B}\right) P\left(E_{B}\right)+P\left(E / E_{W}\right) P\left(E_{W}\right)}$ <br/><br/> Here, <br/><br/> $P\left(E_{R}\right)=3 / 10, \quad P\left(E_{B}\right)=4 / 10, \quad P\left(E_{W}\right)=3 / 10$ <br/><br/> and <br/><br/> $$ \begin{aligned} & P\left(E / E_{R}\right)=5 / 10, \quad P\left(E / E_{B}\right)=6 / 10, \quad P\left(E / E_{W}\right)=5 / 10 \\\\ & \therefore \quad P\left(E_{R} / E\right)=\frac{15 / 100}{15 / 100+24 / 100+15 / 100} \\\\ & =\frac{15}{54}=\frac{5}{18} \end{aligned} $$