4. A box contains 12 balls out of which x are black. If one ball is drawn at random from the box, what is the probability that it will be a black ball? If 6 more black balls are put in the box, the probability of drawing a black ball is now double of what it was before. Find x.

Total number of balls in the bag = 12

Number of black balls in the bag = $x$

$\therefore P(getting\ a\ black\ ball) = \frac{x}{12}$

According to the question,

6 more black balls are added to the bag.

$\therefore$ Total number of balls = $12 + 6 = 18$

And, the new number of black balls = $x+ 6$

$\therefore P'(getting\ a\ black\ ball) = \frac{x+6}{18}$

Also, $P' = 2\times P$

$\implies \frac{x+6}{18} = 2\left (\frac{x}{12} \right )$

$\\ \implies \frac{x+6}{18} = \frac{x}{6} \\ \implies x+6 = 3x \\ \implies 2x = 6$

$\implies x =3$

The required value of $x$is 3

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