Question
Quantity I: A bag contains 5 red, 7 blue, and 8 green
balls. Two balls are drawn randomly. What is the probability that both balls drawn are green? Quantity II: A box contains 4 red, 6 white, and 10 blue balls. If three balls are drawn at random, what is the probability that all of them are blue? In the question, two Quantities I and II are given. You have to solve both the Quantity to establish the correct relation between Quantity-I and Quantity-II and choose the correct option.Solution
Quantity I: Total balls = 5 + 7 + 8 = 20 Favorable outcomes for two green balls = 8C2 = (8 * 7) / (2 * 1) = 28 Total outcomes for drawing two balls = 20C2 = (20 * 19) / (2 * 1) = 190 Probability = 28 / 190 = 0.147 Quantity II: Total balls = 4 + 6 + 10 = 20 Favorable outcomes for three blue balls = 10C3 = (10 * 9 * 8) / (3 * 2 * 1) = 120 Total outcomes for drawing three balls = 20C3 = (20 * 19 * 18) / (3 * 2 * 1) = 1140 Probability = 120 / 1140 = 0.105 Answer: E (Quantity I > Quantity II)
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