CMAT 2016 Solved Question Paper

(c) P$\left({\mathit{E}}_1\right)$ = Probability that bag 1 is chosen = $\frac{1}{2}$
P$\left({\mathit{E}}_2\right)$ = Probability that Bag 2 is chosen = $\frac{1}{2}$
Let P (E) = Probability of getting blue ball
Then, P$\left(\frac{\mathit{E}}{{\mathit{E}}_1}\right)$ = Probability of getting blue ball from bag 1
= $\frac{2}{5}$
∴P $\left(\frac{\mathit{E}}{{\mathit{E}}_2}\right)$ = Probability of getting blue ball from bag 2 = $\frac{4}{5}$
By Bays' theorem, the required probability
P $\left(\frac{{\mathit{E}}_2}{\mathit{E}}\right)$ = $\frac{\mathit{P}\left({\mathit{E}}_2\right).\mathit{P}\left(\frac{\mathit{E}}{{\mathit{E}}_2}\right)}{\mathit{P}\left({\mathit{E}}_1\right).\mathit{P}\left(\frac{\mathit{E}}{{\mathit{E}}_1}\right)+\mathit{P}\left({\mathit{E}}_2\right).\mathit{P}\left(\frac{\mathit{E}}{{\mathit{E}}_2}\right)}$
= $\frac{\frac{1}{2}.\frac{4}{5}}{\frac{1}{2}.\frac{2}{5}+\frac{1}{2}.\frac{4}{5}}$ = $\frac{4}{6}$ = $\frac{2}{3}$