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) 2/27  C) 2/21  D) 2/89  E) 2/11
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?