Question

A box contains 12 balls of which some are red in colour. If 6 more red balls are put in
the box and a ball is drawn at random, the probability of drawing a red ball doubles
than what it was before. Find the number of red balls in the bag.

Answer 1

Let number of red balls be = x

P(red ball) = x/12

If 6 more red balls are added:

The number of red balls = x + 6

P(red ball) = (x+6)/18

Since, (x+6)/18 = 2(x/12)

x = 3

Therefore, there are 3 red balls in the bag.

Answer 2

Step-by-step explanation:

AnswEr:

Given

Total Number of Balls in box = 12

Let us Consider that Red ball be $\bold\red{x}.$

$\dag\:\:\small{\underline{\boxed{\sf{\purple{Probability\:=\:\dfrac{Number\:of \; Favourable\; Outcomes}{Total\; Number\;of \; Outcomes}}}}}}$

Probability of red balls drawing -

$\implies\sf P_{1} = \dfrac{x}{12}$

$\bold{\underline{\sf{According\:to\: Question}}}$

If 6 more red balls are put in the box,

Then, Total Number of Balls = 12 + 6

$\implies\sf\red{ 18 \: Balls}$

Total Number of Red Balls = x + 6

$\implies\sf P_{2} = \dfrac{x + 6}{18}$

Probability of Drawing Red Balls doubles than what it was before. [Given]

So,

$\implies\sf \dfrac{x + 6}{18} = \cancel{2} \times \dfrac{x}{\cancel{12}}$

$\implies\sf \dfrac{x +6}{18} = \dfrac{x}{6}$

$\implies\sf\cancel 6 \times \bigg( \dfrac{x +6}{\cancel{18}}\bigg) = x$

$\implies\sf \dfrac{x +6}{3} = x$

$\implies\sf x + 6 = 3x$

$\implies\sf 6 = 3x - x$

$\implies\sf 6 = 2x$

$\implies\sf x = \cancel\dfrac{6}{2}$

$\implies\large\boxed{\sf{\purple{x\:=\;3}}}$

$\rule{200}2$

$\small\bold{\underline{\sf{\red{Hence,\;There\; are\:3\;Red\;Balls\: in \;the \;bag.}}}}$