UPSC Prelims 2023·CSAT·Quantitative Aptitude·Combinatorics and Probability

A box contains 14 black balls, 20 blue balls, 26 green balls, 28 yellow balls, 38 red balls and 54 white balls. Consider the following statements: 1. The smallest number n such that any n balls drawn from the box randomly must contain one full group of at least one colour is 175. 2. The smallest number m such that any m balls drawn from the box randomly must contain at least one ball of each colour is 167. Which of the above statements is/are correct?

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  1. A1 only
  2. B2 only
  3. CBoth 1 and 2Correct
  4. DNeither 1 nor 2

Explanation

To solve this problem, we use the Worst Case Scenario approach. Total balls: 14 Black, 20 Blue, 26 Green, 28 Yellow, 38 Red, and 54 White. Total sum is 180. Statement 1: To find n such that we must have one full group of at least one colour, we consider the worst case where we are just one ball short of completing every single group. In this case, we draw: 13 Black, 19 Blue, 25 Green, 27 Yellow, 37 Red, and 53 White. Sum = 13 + 19 + 25 + 27 + 37 + 53 = 174 balls. The very next ball (the 175th ball) must complete one of the colour groups. Thus, n is 175. Statement 1 is correct. Statement 2: To find m such that we must have at least one ball of each colour, we consider the worst case where we pick every single ball of the five largest groups first, but none of the smallest group. The groups in descending order are: 54 White, 38 Red, 28 Yellow, 26 Green, 20 Blue, and 14 Black. Sum of the five largest groups = 54 + 38 + 28 + 26 + 20 = 166 balls. At this point, we have 166 balls but still no Black ball. The 167th ball must be Black, ensuring we have at least one ball of every colour. Thus, m is 167. Statement 2 is correct. Since both statements are correct, the answer is C.
Quantitative Aptitude: A box contains 14 black balls, 20 blue balls, 26 green balls, 28 yellow balls, 38 red balls and 54 white balls. Consider

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