something blue balls and 9 red balls

there are 9 red balls
number of blue balls = x

therefore total number of balls = 9+x

probability that the first ball drawn is a red ball = 9/(9+x)

probability that the second ball is also red = 8/(8+x)

probability that the first two balls drawn are red = 6/11 (given)

Therefore, [9/(9+x)]*[8/(8+x)] = 6/11
OR
11*9*8=6*(9+x)(8+x)
11*9*8/6 = 72 +17x + x^2
OR
x^2 + 17x +72 - 132 = 0
x^2 + 17x - 60 = 0
x^2 +20x - 3x - 60 = 0
x(x+20) - 3(x+20) = 0
(x-3)(x+20) = 0

x = 3 or -20

x can't be negative therefore x = 3
i.e. number of blue balls = 3

Hence, the probability that the third ball is a blue ball is 3/(8+3)
or
3/11 (i think if my math was correct)

Hope this helps !!

OA is 9/55