A bag contains 'x' red balls '2x' white balls and '3x' black balls. 3 balls are drawn at random. The probability that all the balls drawn are of different colors is 0.3 .How many white balls are present in the bag?

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Solution

The correct option is A 2
P(All different)= $(^{x} C_{1}) \times (^{2 x} C_{1}) \times (^{3 x} C_{1}) / (^{6 x} C_{3})$

P(E)= $(6 x^{3}) \times \frac{6}{6 x} \times [(6 x - 1) \times (6 x - 2)] = \frac{3}{10}$

or, $\frac{3 x^{2}}{[(6 x - 1) \times (6 x - 2)]} = \frac{3}{10}$

$10 x^{2} = 18 x^{2} - 9 x + 1$

or, (x-1)(8x-1)=0

x can not be a fraction hence x=1

Total number of white balls= 2x=2

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A bag contains 'x' red balls '2x' white balls and '3x' black balls. 3 balls are drawn at random. The probability that all the balls drawn are of different colors is 0.3 .How many white balls are present in the bag?

The correct option is A 2P(All different)= (xC1)×(2xC1)×(3xC1)/(6xC3) P(E)= (6x3)×66x×[(6x−1)×(6x−2)]=310 or, 3x2[(6x−1)×(6x−2)]=310 10x2=18x2−9x+1 or, (x-1)(8x-1)=0 x can not be a fraction hence x=1 Total number of white balls= 2x=2