A bag contains $^{'} x^{'}$ red balls $^{'} 2 x^{'}$ white balls and $^{'} 3 x^{'}$ 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) $= (^{x} C_{1}) \times (^{2 x} C_{1}) \times \frac{(^{3 x} C_{1})}{(^{6 x} C_{3})}$

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

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

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

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

x can not be a fraction hence $x = 1$

Total number of white balls $= 2 x = 2$

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