UPSC NDA II 2024 Mathematics Paper

Concept:Probability of drawing three balls of the same colour with replacement involves independent draws. The probability for each colour is the cube of its individual probability, and the total probability is the sum of these.Explanation:Total balls = $2 + 4 + 6 = 12$.Probability of drawing one black ball = $\frac{2}{12} = \frac{1}{6}$.Probability of drawing three black balls with replacement = $\left(\frac{2}{12}\right)^3 = \frac{2^3}{12^3} = \frac{8}{1728}$.Probability of drawing one yellow ball = $\frac{4}{12} = \frac{1}{3}$.Probability of three yellow balls = $\left(\frac{4}{12}\right)^3 = \frac{4^3}{12^3} = \frac{64}{1728}$.Probability of drawing one white ball = $\frac{6}{12} = \frac{1}{2}$.Probability of three white balls = $\left(\frac{6}{12}\right)^3 = \frac{6^3}{12^3} = \frac{216}{1728}$.Since the events are mutually exclusive, total probability = $\frac{8}{1728} + \frac{64}{1728} + \frac{216}{1728} = \frac{288}{1728} = \frac{1}{6}$.Answer:Option A: $\frac{1}{6}$.