Out of nine balls three balls can be chosen =9C3 ways ∴ Sample space =9C3=
9!
3!6!
=
9×8×7
6
=84 According to the question, all three ball should be different. So out of 3 red balls 1 is chosen and out of 4 blue 1 is chosen and out of 2 green 1 is chosen. ∴ Total cases =3C1×4C1×2C1=3×4×2=24 ∴ Probability =
