A bag contains 30 red jellybeans and 30 blue jellybeans. If 3 jellybeans are removed one at a time, at random and are not replaced, what is the probability that all 3 jellybeans removed from the bag are blue? Choices
A) 9/100  B) 2/19  C) 1/8  D) 1/20  E) 3/10
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 = 20C3 = 1140   Favourable Outcomes = No. of ways to choose 3 Jelly bean such that they are all Blue out of 10 Blue = 10C3 = 120   Probability = Favourable Outcomes / Total Outcomes = 10C3 / 20C3   Probability = 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) = 1/20   Answer: option 1/20
<result>D</result>
A bag contains 30 red jellybeans and 30 blue jellybeans. If 3 jellybeans are removed one at a time, at random and are not replaced, what is the probability that all 3 jellybeans removed from the bag are blue?