A bag contains 10 red jellybeans and 10 blue jellybeans. If 3 jellybeans are removed one at a time, at random and are not replaced, what is the probability E that all 3 jellybeans removed from the bag are blue? Choices:	A) 9 / 100	B) 2 / 19	C) 1 / 8	D) 3 / 20	E) 3 / 10

Rationale: "method - 1 10 red jellybeans and 10 blue jellybeans total outcomes = no . of ways to choose 3 jelly bean at random out of a total 20 jellybeans = 20 c 3 = 1140 favourable outcomes = no . of ways to choose 3 jelly bean such that they are all blue out of 10 blue = 10 c 3 = 120 probability = favourable outcomes / total outcomes = 10 c 3 / 20 c 3 probability e = 120 / 1140 = 2 / 19 answer : option b method - 2 probability of first jelly bean to be blue = 10 / 20 [ total 10 blue out of total 20 jellybeans ] probability of second jelly bean to be blue = 9 / 19 [ total 9 blue remaining out of total 19 jellybeans remaining ] probability of third jelly bean to be blue = 8 / 18 [ total 8 blue remaining out of total 18 jellybeans remaining ] required probability = ( 10 / 20 ) * ( 9 / 19 ) * ( 8 / 18 ) = 2 / 19 answer : option b"
Answer: B