**Answer: (c) \(\frac{10}{19}\)**

**Let the events E**_{1,} E_{2,} and A be defined as follows:

E_{1} = Choosing bag A

E_{2} = Choosing bag B

A = Choosing red ball.

Then, **P(E**_{1}) = P(E_{2}) = \(\frac{1}{2}\)

(∵ There are **two bags that have an equally likely chance of being chosen**)

P(A/E_{1}) = P(Drawing red ball from bag A) = \(\frac{2}{4} =\frac{1}{2}\) **(∵ 2 red out of 4 balls) **

P(A/E_{2}) = P(Drawing red ball from bag B) = \(\frac{5}{9}\) **(5 red out of 9 balls)**

∴ P(Red balls are drawn from bag B)

\(=P(E_2/A) = \frac{P(E_2)\times P(A/E_2)}{P(E_1)\times P(A/E_1)+P(E_2)\times P(A/E_2)}\)

\(=\frac{\frac{1}{2}\times \frac{5}{9}}{\frac{1}{2}\times \frac{1}{2}+ \frac{1}{2}\times \frac{5}{9}}\) = \(\frac{\frac{5}{18}}{\frac{1}{4}+\frac{5}{18}}\) = \(=\frac{5/18}{19/36}\) = **\(\frac{10}{19}\)**