Question

Bag I contains 1 white, 2 black and 3 red balls; Bag II contains 2 white, 1 black and 1 red balls; Bag III contains 4 white, 3 black and 2 red balls. A bag is chosen at random and two balls are drawn from it with replacement. They happen to be one white and one red. What is the probability that they came from Bag III.

Answer 1

Bag I- 1 W, 2 B, 3 R

Bag II- 2 W, 1 B, 1 R

Bag III= 4 W, 3 B, 2 R

Probability that first bag is chosen and one is white and other red=$\frac{1}{3}\ast \frac{1\ast 3}{6\ast 6}$

Probability that second bag is chosen and one is white and other red=$\frac{1}{3}\ast \frac{1\ast 2}{4\ast 4}$

Probability that thirdbag is chosen and one is white and other red=$\frac{1}{3}\ast \frac{4\ast 2}{9\ast 9}$

using Bayes rule

Probability that it came from third bag=$\frac{\frac{1}{3}\ast \frac{4\ast 2}{9\ast 9}}{\frac{1}{3}\ast \frac{1\ast 3}{6\ast 6}+\frac{1}{3}\ast \frac{1\ast 2}{4\ast 4}+\frac{1}{3}\ast \frac{4\ast 2}{9\ast 9}}=\frac{64}{199}$