There are 12 balls ¡n a bag in which x are white(i) What is the proba

1.	There are 12 balls ¡n a bag in which x are white(i) What is the probability that if ball drawn at random is white.(ii) If 6 more white balls are put, (i) the probability of the ball drawn is white becomes double, find the value of x.
Answer» (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

There are 12 balls ¡n a bag in which x are white(i) What is the probability that if ball drawn at random is white.(ii) If 6 more white balls are put, (i) the probability of the ball drawn is white becomes double, find the value of x.

Answer»

(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

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