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UPSC CSE CSAT 2023 · Paper II (CSAT) · Question 36 · Counting, permutations & probability

किसी डिब्बे में 14 काली गेंदें, 20 नीली गेंदें, 26 हरी गेंदें, 28 पीली गेंदें, 38 लाल गेंदें और 54…

UPSC Civil Services (Preliminary) Examination 2023, General Studies Paper II (CSAT), Question 36 — Counting, permutations & probability.

CSAT 2023 · Q36

Counting, permutations & probability Medium

किसी डिब्बे में 14 काली गेंदें, 20 नीली गेंदें, 26 हरी गेंदें, 28 पीली गेंदें, 38 लाल गेंदें और 54 सफ़ेद गेंदें हैं । निम्नलिखित कथनों पर विचार कीजिए :

  1. 1.यदि डिब्बे में से यादृच्छिक रूप से कोई सी n गेंदें इस प्रकार निकाली जाएँ कि उनमें कम-से-कम एक रंग का एक पूरा समूह अवश्य हो, तो n की लघुतम संख्या 175 है ।
  2. 2.यदि डिब्बे में से यादृच्छिक रूप से कोई सी m गेंदें इस प्रकार निकाली जाएँ कि उनमें हर रंग की कम-से-कम एक गेंद अवश्य हो, तो m की लघुतम संख्या 167 है ।

उपर्युक्त कथनों में से कौन-सा/से सही है/हैं ?

Read this question in English

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?

(a) 1 only · (b) 2 only · (c) Both 1 and 2 · (d) Neither 1 nor 2

उत्तर और व्याख्या · explanation in English

Answer: (c) 1 और 2 दोनों

For a full group, the unluckiest draw takes every colour one ball short of complete: 13 + 19 + 25 + 27 + 37 + 53 = 174, so 175 balls guarantee a full group. For one ball of each colour, the unluckiest draw takes every ball except the smallest colour: 180 − 14 = 166, so 167 balls guarantee all six colours.

  1. Total balls = 14 + 20 + 26 + 28 + 38 + 54 = 180.
  2. Statement 1: the worst case leaves every colour one short: 13 + 19 + 25 + 27 + 37 + 53 = 174 balls with no full group. The next ball completes one, so n = 175. Correct.
  3. Statement 2: the worst case is all balls except the smallest colour (14 black): 180 − 14 = 166 balls without black. The next ball must be black, so m = 167. Correct.
  • ✓ 1. 174 balls can still leave every colour one ball short of full; the 175th ball must complete a group.
  • ✓ 2. 166 balls can all be non-black; the 167th must be black, giving all six colours.

Remember · Guarantee questions: build the unluckiest draw that still fails, then add one.

Question and answer: UPSC's official GS Paper II (2023, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). ·

Cite this question: UPSC CSE CSAT 2023, Q36 — minimalistias.com/pyqs/csat/2023/q36

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