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?	Select
A) 4/8
B) 9/2
C) 7/6
D) 9
E) 2/21
Explanation:   Given that there are three blue marbles, four red marbles, six green marbles and two yellow marbles. Probability that both marbles are blue = Â³Câ‚‚/Â¹âµCâ‚‚ = (3 * 2)/(15 * 14) = 1/35   Probability that both are yellow = Â²Câ‚‚/Â¹âµCâ‚‚ = (2 * 1)/(15 * 14) = 1/105   Probability that one blue and other is yellow = (Â³Câ‚ * Â²Câ‚)/Â¹âµCâ‚‚ = (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>E</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?