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Bag a Contains 1 White, 2 Blue and 3 Red Balls. Bag B Contains 3 White, 3 Blue and 2 Red Balls. - Mathematics

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Sum

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. 

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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 B1, B2, B3 and E be the events defined as
B1: Bag A is selected
B2: Bag B is selected
B3: Bag C is selected
And E: 1 white and 1 red ball is drawn 

P(B1) = P(B2) = P(B3) = `1/3`

P(E/B1) = `(""^1C_1 xx^3C_1)/(""^6C_2) = 1/5`

P(E/B2) = `(""^3C_1 xx ^2C_1)/(""^8C_2) = 3/14`

P(E/B3) = `(""^2C_1 xx ^4C_1)/(""^9C_2) = 2/9`

Required probability = P(B1) × P(E/B1) + P(B2) × P(E/B2) + P(B3)  × P(E/B3)

`= 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`

Concept: Introduction of Probability
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