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?

Initially, total number of balls = 12
No. of black balls = x
So, P(black ball) = x/12
If 6 more black balls are added, total number of balls = 12 + 6 = 18
No. of black balls = x + 6
P(black ball) = (x + 6)/18
Given the probability of drawing a black ball is now double of what it was before
So, 2 * (x/12) = (x + 6)/18
(2x)/12 = (x + 6)/18
x/6 = (x + 6)/18
3x = x + 6
2x = 6
x = 3
So, the number of black balls present initially = 3.