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?
- (a) 1/2
- (b) 4/7 ✓ UPSC's answer
- (c) 5/9
- (d) 12/13
Why the answer is (b)
• Total balls = 15 red + 20 black = 35.
• Red balls numbered 3: 40% of 15 = 6.
• Black balls numbered 1 or 2: Since 45% are numbered 2 and 30% are numbered 3, the remaining 25% are numbered 1. So, black balls numbered 1 or 2 = 45% + 25% = 70% of 20 = 14.
• Total winning balls = 6 (Red, No. 3) + 14 (Black, No. 1 or 2) = 20.
• Probability of winning = 20/35 = 4/7.
Why the other options are wrong
- (a) 1/2
- 1/2 implies 17.5 winning balls, which is not an integer and does not match the calculated 20.
- (c) 5/9
- 5/9 implies approximately 19.44 winning balls, which does not match the calculated 20.
- (d) 12/13
- 12/13 implies approximately 32.3 winning balls, which is far higher than the calculated 20.
Asked in the CSAT Paper II of the UPSC Civil Services Preliminary Examination 2018, held on 3 June 2018. Question and answer key: Union Public Service Commission. Explanation: UPSC Answer Check.