Question

A bag A contains 2 white and 2 red balls and another bag B contains 4 white and 5 red balls. A ball is drawn and is found to be red.The probability that it was drawn from bag B is: (a) \(\frac{5}{19}\) (b) \(\frac{21}{52}\) (c) \(\frac{10}{19}\) (d) \(\frac{25}{52}\) Select the correct answer from above options

Answer 1

Answer: (c) \(\frac{10}{19}\) Let the events E1, E2, and A be defined as follows: E1 = Choosing bag A E2 = Choosing bag B A = Choosing red ball. Then, P(E1) = P(E2) = \(\frac{1}{2}\) (∵ There are two bags that have an equally likely chance of being chosen) P(A/E1) = P(Drawing red ball from bag A) = \(\frac{2}{4} =\frac{1}{2}\) (∵ 2 red out of 4 balls) P(A/E2) = P(Drawing red ball from bag B) = \(\frac{5}{9}\) (5 red out of 9 balls) ∴ P(Red balls are drawn from bag B) \(=P(E_2/A) = \frac{P(E_2)\times P(A/E_2)}{P(E_1)\times P(A/E_1)+P(E_2)\times P(A/E_2)}\) \(=\frac{\frac{1}{2}\times \frac{5}{9}}{\frac{1}{2}\times \frac{1}{2}+ \frac{1}{2}\times \frac{5}{9}}\) = \(\frac{\frac{5}{18}}{\frac{1}{4}+\frac{5}{18}}\) = \(=\frac{5/18}{19/36}\) = \(\frac{10}{19}\)