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Two balls are drawn at random from a bag containing 3 white, 3 red, 4 green and 4 black balls, one by one without replacement. Find the probability that both the balls are of different colours. |
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Answer» 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.
Since all cases are mutually exclusive, therefore Reqd. Probability = \(\frac{3}{14}\) x \(\frac{11}{13}\) + \(\frac{3}{14}\) x \(\frac{11}{13}\) + \(\frac{4}{14}\) x \(\frac{10}{13}\) + \(\frac{4}{14}\) x \(\frac{10}{13}\) = \(\frac{33+33+40+40}{14\times13}\) = \(\frac{146}{182}\) = \(\frac{73}{91}.\) |
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