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There are 3 bags containing 3 colored balls - Black, Green and White. Bag 1 contains: 15 black balls. Chances of drawing a white ball is 2/5. The ratio of number of green ball to the number of white ball is 3:4. Bag 2 contains: Number of black balls and number of white balls are equal. Number of green ball is 4/5th of the number of green ball in bag 1. The Chances of black ball is 5/14. Bag 3 contains: Number of Black Balls is (1/3)rd of the total number of black balls in Bag 1 and Bag 2 together. Number of Green ball is half of the number of white ball in bag 1. The probability of drawing a white ball is 3/7.

If one ball is drawn from bag 1 and 1 ball is drawn from bag 3. Find the Chances the these are green ball.

  1. 3/35

  2. 9/245

  3. 11/245

  4. 4/35

  5. 8/35
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A Correct answer
Explanation

Bag 1: White = 2/5, Black = 15. If White = 2/5, then Black+Green = 3/5. Ratio G:W = 3:4. Let G=3x, W=4x. Total = 7x. W/(7x) = 4/7? No, the problem says P(W)=2/5. Let total balls in Bag 1 be N1. W/N1 = 2/5. G/W = 3/4 => G = 3/4 W. B=15. N1 = 15 + W + 3/4 W = 15 + 1.75W. W/(15+1.75W) = 0.4 => W = 6 + 0.7W => 0.3W = 6 => W=20. G=15. Total=50. P(G1) = 15/50 = 3/10. Bag 3: P(W)=3/7. G = 1/2 W(bag1) = 10. B = 1/3(B1+B2). B1=15. B2: B=W, G=4/5*G1=12. P(B)=5/14. B/(B+W+12)=5/14. B/(2B+12)=5/14 => 14B = 10B + 60 => 4B=60 => B=15. B2=15, W2=15. B3 = 1/3(15+15) = 10. Total Bag 3 = 10(B)+10(G)+W3. W3/Total = 3/7. W3/(20+W3) = 3/7 => 7W3 = 60 + 3W3 => 4W3=60 => W3=15. Total Bag 3 = 35. P(G3) = 10/35 = 2/7. P(G1 and G3) = 3/10 * 2/7 = 6/70 = 3/35.

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