If three marbles are picked at random, what is the probability that two are blue and one is yellow?

Ways of selection of two blue marbles = ⁴C₂
Ways of selection of one yellow marble = ³C₁
Ways of selection of three marbles = ¹⁵C₃
So, required probability = (⁴C₂ x ³C₁) / (¹⁵C₃)
= [4!/{2! (4 - 2)!} x 3!/{1! (3 - 1)!}] / [15! / {3! (15 - 3)!}]
= [(4 x 3 x 2 x 1) / (2 x 1 x 2 x 1)] x [(3 x 2 x 1) / (1 x 2 x 1)] / [ (15 x 14 x 13 x 12!) / (3 x 2 x 12!)]
= (6 x 3) / (5 x 7 x 13)
= 18 / 455