A focus group is currently made up of x red balls and y green balls. If 5 red balls and 4 green balls are added to the group, and if one ball is selected at random from the larger focus group, then what is the probability that a green ball is selected?	Choose the correct choice:
A) y/x+5
B) y/x+4
C) (y+5)/(x+y+9)
D) (y+4)/(x+y+9)
E) (x+4)/(x+y+9)
x red balls   y green balls   x+4+y+5 is the total   =x+y+9   y+4 is the number of green balls   so (y + 4)/(x + y + 9)   the answer is (y+4)/(x+y+9)
<result>D</result>
A focus group is currently made up of x red balls and y green balls. If 5 red balls and 4 green balls are added to the group, and if one ball is selected at random from the larger focus group, then what is the probability that a green ball is selected?