Total No. of balls = 4 + 5 + 6 = 15
n(S) = ^{15}C_{3} = 15x14x13/3x2 = 455
2 green balls can be selected from 5 green balls in ^{5}C_{2} ways and the remaining one ball can be selected from the remaining 15  5 = 10 balls
in ^{10}C_{1} ways.
n(E) = ^{5}C_{2} x ^{10}C_{1} = 10 x 10 = 100
∴ P(E) = 100/455 = 20/91
