Total number of balls = 2 + 3 + 4 = 9
Probability to get 2 balls = 9C2

Probability to get 2 Red balls = 2C2
Probability to get 2 White balls = 3C2
Probability to get 2 Black balls = 4C2

Probability to get 2 balls of same color = (2C2 + 3C2 + 4C2)/ 9C2 = 5/18

I would think that you want the sum of the following probabilities:

a) P (2 red balls)
b) P (2 white balls)
c) P (2 black balls)

If I'm correct about that, then...

a) P (2 red balls) = (2/9) * (1/8) = 2/72
b) P (2 white balls) = (3/9) * (2/8) = 6/72
c) P (2 black balls) = (4/9) * (3/8) = 12/72

Summing, (2/72) + (6/72) + (12/72) = 20/72 = 5/18

Of course, if my initial assumption is incorrect, then ignore everything.

do you count probability or the number of subsets?
below 9C2 should return the total number of subsets, it's not probability
if the problem asked not *without replacement* the method of Tomada would be the best one. However, if we approach with the method of MakeYourTimeCount here we must consider this only for *the same color* balls selected *without replacement*.

@nikhilgupta, to avoid confusion in test use method of Tomada
use the method of MakeYourTimeCount to improve your strategy with combinatorics question types

MakeUrTimeCount wrote:Total number of balls = 2 + 3 + 4 = 9
Probability to get 2 balls = 9C2

Probability to get 2 Red balls = 2C2
Probability to get 2 White balls = 3C2
Probability to get 2 Black balls = 4C2

Probability to get 2 balls of same color = (2C2 + 3C2 + 4C2)/ 9C2 = 5/18