Given, 3 white (3 W), 3 red (3 R), 4 green (4 G), 4 black (4 B) balls Total no. of balls = 3 + 3 + 4 + 4 = 14 Two balls are to be drawn, one by one without replacement. There are 4 possibilities.First BallSecond BallCase IWhiteNot whiteCase IIRedNot RedCase IIIGreenNot GreenCase IVBlackNot blackSince all cases are mutually exclusive, thereforeReqd. Probability = \(rac{3}{14}\) x \(rac{11}{13}\) + \(rac{3}{14}\) x \(rac{11}{13}\) + \(rac{4}{14}\) x \(rac{10}{13}\) + \(rac{4}{14}\) x \(rac{10}{13}\) = \(rac{33+33+40+40}{14 imes13}\) = \(rac{146}{182}\) = \(rac{73}{91}.\)