A bag contains 15 red balls and 20 black balls. Each ball is numbered either 1 or 2 or 3. 20% of the red balls are numbered 1 and 40% of them are numbered 3. Similarly, among the black balls, 45% are numbered 2 and 30% are numbered 3. A boy picks a ball at random. He wins if the ball is red and numbered 3 or if it is black and numbered 1 or 2. What are the chances of his winning?
Answer & explanation
Answer: (b) 4/7
Winning balls are the red ones numbered 3 (40% of 15 = 6) and the black ones numbered 1 or 2 (70% of 20 = 14). That is 20 winning balls out of 35, a chance of 4/7.
- Red numbered 3: 40% of 15 = 6.
- Black numbered 3: 30% of 20 = 6, so black numbered 1 or 2 = 20 − 6 = 14.
- Winning balls = 6 + 14 = 20 out of 35.
- Probability = 20/35 = 4/7.
Remember · Count the favourable balls directly; use a complement (100% − 30%) when it is quicker than adding separate percentages.
Question and answer: UPSC's official GS Paper II (2018, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). ·