UPSC Prelims 2018 · CSAT
Question 39: Basic Numeracy
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?
AnswerUPSC’s answer is (b).
Red balls: 15, of which 20% = 3 carry number 1 and 40% = 6 carry number 3. Black balls: 20, of which 45% = 9 carry number 2 and 30% = 6 carry number 3, so 5 carry number 1. He wins with a red 3 (6 balls) or a black 1 or 2 (5 + 9 = 14 balls). That is 20 winning balls out of 35, so the probability is 20/35 = 4/7.
Why each option is right or wrong
- (a) 1/2 would need 17.5 winning balls out of 35.
- (b) Correct: 20 winning balls out of 35 is 4/7.
- (c) 5/9 does not give a whole number of balls out of 35.
- (d) 12/13 does not fit, since 35 is not a multiple of 13.
Answer: UPSC’s key. Explanation: written by UPSC Grade, not by UPSC.
Source: UPSC, Civil Services (Preliminary) Examination 2018, General Studies Paper II (CSAT), Series A. The paper on upsc.gov.in.