Question

A bag contains 15 red balls and 20 black balls. Each ball is numbered either 1or 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) \(\frac 12\)

(b) \(\frac 47\)

(c) \(\frac 59\)

(d) \(\frac{12}{13}\)

Answer 1

Correct option is (b) \(\frac 47\)

The winning condition is if the ball is red and numbered 3 or if it is black and numbered 1 and 2

Total number of balls in bag = 15 + 20 = 35

Number of red balls numbered 3 = 40% of 15 = 6

Number of black balls numbered 1 or 2 = 70% of 20 = 14

The probability of winning = \(\frac{6+14}{35}\) = \(\frac 47\)