(i) The number of balls in the bag = 12
∴ The number of all possible outcomes = 12
The number of white balls = x
∴ The number of favourable outcomes = x
∴ Required probability = \(\frac { x }{ 12 }\)
(ii) After putting 6 more white balls in the bag, the total number of ball
= 12 + 6 = 18
The number of white balls = (x + 6)
The new probability = \(\frac { x+6 }{ 18 }\)
According to the given problem,
\(\frac { x+6 }{ 18 }\) = \(\frac { x }{ 12 }\) × 2
⇒ \(\frac { x+6 }{ 18 }\) = \(\frac { x }{ 6 }\)
⇒ 6x + 36 = 18x
⇒ 18x – 6x = 36
⇒ 12x = 6
⇒ x = \(\frac { 36 }{ 12 }\) = 3