Question

Aboxcontains12ballsofwhichsomeareredincolour.If6moreredballsareput intheboxandaballisdrawnatrandom,theprobabilityofdrawingaredballdoubles thanwhatitwasbefore.Findthenumberofredballsinthebag.

Answer 1

let no. red balls be x
total balls are 12
6 red balls are added
P(getting a red ball earlier)=x/12
P(getting a red ball after adding 6 red balls)=(6+x)/(12+8)
$\frac{x + 6}{18} = 2 (\frac{x}{12}) \\ \frac{x + 6}{18} = \frac{x}{6} \\ 6(x + 6) = 18x \\ 6x + 36 = 18x \\ 18x - 6x = 36 \\ 12x = 36 \\ x = 3$
no. of red balls are 3 (earlier)

Answer 2

Total number of balls = 12
Total number of red balls = x
Probability of red ball = x/12
Total number of balls when 6 balls are added = 18
Now probability of red balls = x+6/18

According to the question,
x+6/18 =2(x/12)
x+6/18 = x/6
6x+36 = 18x
12x = 36
x = 3
therefore the number of red balls in bag is 3.

Hope this helps you...:)