The probability of picking 2 red balls one after another without repla

Probability of picking 2 red balls = 14/33
(R/R+W)*(R-1)/(R+W-1) = 14/33 = (2*7)/(3*11) = (2/3)*(7/11) --> Once the value is reduced to this level, it should be easy to find out the number of red balls.

1) Considering the above expression, both numerator and denominator must be consecutive.
2) Lets multiply both numerator and denominator with the same integer so that the value remains unaffected.
3) We can't multiply 7/11 as we need a consecutive term in both numerator and denominator. So lets multiply (2/3) to make the numerator and denominator consecutive.

(2/3)*(2/2) = 4/6 --> 4*7/6*11 --> Numerator and denominator not consecutive
(2/3)*(3/3) = 6/9 --> 6*7/9*11 --> Only numerator is consecutive
(2/3)*(4/4) = 8/12 --> 8*7/12*11 --> Numerator and denominator are consecutive here

Number of red balls, R = 8.
Total number of balls, R + W = 12 (This value is < 20, satisfying the condition)
Number of white balls, W = 12 - 8 = 4

Probability of picking 2 white balls w/o replacement = (W/R+W)*(W-1)/(R+W-1) = (4/12)*(3/11) = 3/33

Answer: B