1.

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.

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.

First BallSecond Ball
Case IWhiteNot white
Case IIRedNot Red
Case IIIGreenNot Green
Case IVBlackNot black

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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