Bag A contains 1 white, 2 blue and 3 red balls. Bag B contains 3 white, 3 blue and 2 red balls. Bag C contains 2 white, 3 blue and 4 red balls. One bag is selected at random and then two balls are drawn from the selected bag. Find the probability that the balls draw n are white and red.

#### Solution

Given:

Bag A: 1 white, 2 blue and 3 red balls

Bag B: 3 white, 3 blue and 2 red balls

Bag C: 2 white, 3 blue and 4 red balls

Let B_{1}, B_{2}, B_{3} and E be the events defined as

B_{1}: Bag A is selected

B_{2}: Bag B is selected

B_{3}: Bag C is selected

And E: 1 white and 1 red ball is drawn

P(B_{1}) = P(B_{2}) = P(B_{3}) = `1/3`

P(E/B_{1}) = `(""^1C_1 xx^3C_1)/(""^6C_2) = 1/5`

P(E/B_{2}) = `(""^3C_1 xx ^2C_1)/(""^8C_2) = 3/14`

P(E/B_{3}) = `(""^2C_1 xx ^4C_1)/(""^9C_2) = 2/9`

Required probability = P(B_{1}) × P(E/B_{1}) + P(B_{2}) × P(E/B_{2}) + P(B_{3}) × P(E/B_{3})

`= 1/3 xx 1/5 + 1/3 xx 3/14 + 1/3 xx 2/9`

`= 1/3 [1/5 + 3/14 + 2/9]`

`= 1/3 [(126 + 135 + 140)/(5xx14xx9)]`

`= 401/(3xx5xx14xx9) = 401/1890`