In a box there are A green balls, 3A + 6 red balls and 2 yellow ones.

I think the information anout number of green balls is missing in the question.

Without that we cannot get the total no of balls and cannot find the probability.

The question just said there are green balls, 3A+2 red and 2 yellow.

Total = green balls + 3A+2+ 2

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Probability space is G + R + Y and we want the probability of G+Y
\(\frac{(G+Y)}{(G+R+Y)}\)

G=A
R=3A+6
Y=2

\(\frac{(A+2)}{(4A+8)}\)

\(\frac{(A+2)}{[4(A+2)]}\)
\(\frac{1}{[4(1)]}=\frac{1}{4}\)

green balls = A

red balls = 3A+6

yellow balls = 2

Total numbers of ball = 4A+8

we have to find the probability of taking out a green or a yellow ball

since both, the events are mutually exclusive (ones that cannot happen at the same time)

probability 0f taking out (a green or a yellow) = probability of taking out (a green) + probability of taking out (a yellow)

probability 0f taking out (a green or a yellow) = \(\frac{green ball}{total number of ball}\) + \(\frac{yellow ball}{total number of ball}\)

probability 0f taking out (a green or a yellow) = \(\frac{A}{4A+8}\)+\(\frac{2}{4A+8}\)

probability 0f taking out (a green or a yellow) = \(\frac{(A+2)}{4(A+2)}\) = \(\frac{1}{4}\)

D is the answer

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