Bag I contains 4 red and 5 black balls, while another Bag II contains 5 red and 6 black balls. One ball is drawn at random from one of the bags and it is found to be black. Then the probability that it was drawn from Bag II, is

\(\frac{54}{109}\)

Let E₁​, E₂​ be the event of choosing bag 1 and bag 2 respectively.

E be the event that the drawn ball is black.

Then, $P(E_1)=P(E_2)=\frac{1}{2}$

$P(E/E_1)=\frac{5}{9}$

$P(E/E_2)=\frac{6}{11}$

By applying Bayes theorem,

$P(E_2/E_1)=\frac{\frac{1}{2}×\frac{6}{11}}{\frac{1}{2}×\frac{5}{9}+\frac{1}{2}×\frac{6}{11}}=\frac{\frac{6}{22}}{\frac{5}{18×11}+\frac{6}{22×9}}=\frac{\frac{6}{22×9}}{\frac{55+54}{198}}=\frac{\frac{54}{198}}{\frac{109}{198}}$

$\frac{54}{198}×\frac{198}{109}=\frac{54}{109}$