Question

A box contains 12 balls of which some are you 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 then what it was before find the number of red balls in the bag

Answer 1

Total no. Of balls = 12 balls

Let the no. Of red balls be = X

Probability of an event =

favourable outcomes / total outcomes

P ( red ball) =

$\frac{x}{12}$

Now, if 6 more red balls are added

Then, Total no. Of balls now = 18

Total number of red balls = X + 6

P( red ball now) =

$\frac{x + 6}{18}$

According to the question,

2 X [P( Red ball then)] = P ( red ball now)

$2 \times \frac{x}{12 } = \frac{x + 6}{18}$

18X = 6x + 36

12x = 36

x = 3

Thus the number of red balls = 3

Answer 2

$\sf\red{Answer :-}$

Total Number of Balls in box = 12

Let us Consider that Red ball be x

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

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,

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

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

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

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

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

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

$\implies\sf 6 = 2x$

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

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

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