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 red or green?  Pick one
A) 53/105  B) 43/105  C) 42/105  D) 32/105  E) 46/105
given that there are three blue marbles, four red marbles, six green marbles and two yellow marbles. Probability that both marbles are red = 4Câ‚‚/Â¹âµCâ‚‚ = (3 * 2)/(15 * 14) = 3/21   Probability that both are green = 6Câ‚‚/Â¹âµCâ‚‚ = (2 * 1)/(15 * 14) = 1/105   Probability that one red and other is green = (4Câ‚ * 6Câ‚)/Â¹âµCâ‚‚ = (2 * 3 * 2)/(15 * 14) = 27/105   Required probability = 3/21 + 1/105 + 27/105   = 43/105   Answer:43/105
<result>B</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 red or green?