Badminton Feather

something about probability. please help.?

A grade 9 class playing badminton in the gym didn't clean up very well after they were done. When the janitor swept up, there were 12 feathered shuttlecocks and 9 plastic shuttlecocks in the pile. They were put in a bag to show to the class the next day. if the 2 shuttlecocks were pulled out of the bag without replacement, determine the probability of the following a) 1 plastic and then one feathered shuttlecock b) 1 plastic and one feathered shuttlecock c) at least one plastic shuttlecock.

(a) (the number of ways of selecting a plastic shuttlecock first)∙(the number of ways of selecting a feathered shuttlecock second)/(the number of ways of selecting from 21)∙(the number of ways of selecting from 20) = (9∙12) / (21∙20) = 9/35.

(b) [((the number of ways of selecting a plastic shuttlecock first)∙(the number of ways of selecting a feathered shuttlecock second))+(the number of ways of selecting a feathered shuttlecock first)∙(the number of ways of selecting a plastic shuttlecock second))]/(the number of ways of selecting from 21)∙(the number of ways of selecting from 20) = [ (9∙12) + (12∙9)] / (21∙20) = 18/35.

(c) P(at least one plastic shuttlecock in 2 picks) = 1 - P(no plastic shuttlecock in 2 picks) = 1 - (the number of ways of selecting two feathered shuttlecocks from 12)/(the number of ways of selecting two shuttlecocks from 21) = 1 - C(12,2)/C21,2) = 1 - [(12∙11)/2]/[(21∙20)/2] = 1 - 11/35 = 24/35. Note that here you can as well think in terms of permutations as C(12,2)/C(21,2) = P(12,2)/P(21,2).