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In a container, there are 6 red marbles, blue marbles are 40% more than red marbles and green marbles are 25% less than blue marbles. Two marbles are drawn without replacement. If the first marble comes out to be green, then 50% of the red marbles are taken out of the container. (Approximating marbles to nearest whole number) Find the probability that first marble is green and second is red?

In a container, there are 6 red marbles, blue marbles are 40% more than red marbles and green marbles are 25% less than blue marbles. Two marbles are drawn without replacement. If the first marble comes out to be green, then 50% of the red marbles are taken out of the container. (Approximating marbles to nearest whole number) Find the probability that first marble is green and second is red? Correct Answer 9/160

Red Marbles = 6

∵ Blue Marbles are 40% more than red marbles

⇒ Blue Marbles = 6 × 1.4 = 8.4 ≈ 8

∵ Green Marbles are 25% less than blue marbles

⇒ Green Marbles = 8 × 0.75 = 6

Total Marbles = 6 + 8 + 6 = 20

Probability of drawing first green marble = 6C1/20C1 = 6/20 = 3/10

As per question, now 50% of the red marbles are taken out

⇒ Red Marbles = 6 × 0.5 = 3

∵ 1 green marble is already drawn out, so only 5 green balls are left

Total Marbles Left = 3 + 8 + 5 = 16

Probability of Drawing second ball that is red = 3C1/16C1 = 3/16

P(first ball is green and second is red) = (3/10) × (3/16)

∴ Probability = 9/160

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