A box contains 3 blue marbles, 4 red, 6 green marbles and 2 yellow marbles. If two marbles are picked at random, what is the probability that they are either blue or yellow?	Choices:
A) 2/28
B) 2/25
C) 2/21
D) 2/08
E) 2/18
Given that there are three blue marbles, four red marbles, six green marbles and two yellow marbles. Probability that both marbles are blue = 32/212/152/212 = (3 * 2)/(15 * 14) = 1/35   Probability that both are yellow = 22/212/152/212 = (2 * 1)/(15 * 14) = 1/105   Probability that one blue and other is yellow = (32/211 * 22/211)/152/212 = (2 * 3 * 2)/(15 * 14) = 2/35   Required probability = 1/35 + 1/105 + 2/35   = 3/35 + 1/105 = 1/35(3 + 1/3)   = 10/(3 * 35) = 2/21   Answer:2/21
<result>C</result>
A box contains 3 blue marbles, 4 red, 6 green marbles and 2 yellow marbles. If two marbles are picked at random, what is the probability that they are either blue or yellow?