Solution:
All out number of debases = x
All out number of balls = 12
P(E) = (Number of positive results/Total number of results)
P (getting debases) = x/12 — — — – (I)
Presently, when 6 more debases are added,
Absolute balls become = 18
∴ Total number of renounces = x+6
Presently, P (getting renounces) = (x+6)/18 — — — – (ii)
It’s given that, the likelihood of drawing a renounce now is twofold of what it was previously
(ii) = 2 × (I)
(x+6)/18 = 2 × (x/12)
x + 6 = 3x
2x = 6
∴ x = 3