+1245 Container A holds 4 red balls and 6 green balls; containers B and C each hold 6 red balls and 4 green balls. A container is selected at random and then a ball is randomly selected from that container. What is the probability that the ball selected is green? Express your answer as a common fraction.
There are three different possibilities for our first decision, each corresponding to which container we choose. So, if we choose container A, with 1/3 probability, we have a 6/10=3/5 probability of drawing green, which means we have a (1/3)(3/5)=1/5 of picking Container A and then picking a green ball. Similarly for container B the probability is (1/3)(4/10)=2/15, and the same for container C. So, the total probability is (1/5)+(2/15)+(2/15) = 7/15.
