A box contains 12 identical balls of which 5 are red, 4 blue, and the rest are green. If a ball is selected at random from the box, what is the probability that it is green?
\(\frac{3}{4}\)
\(\frac{1}{2}\)
\(\frac{1}{3}\)
\(\frac{1}{4}\)
Correct answer is D
No. of red = 5
No. of blue = 4
No. of green = 3
Total = 12
Pr(Green) = \(\frac{3}{12} = \frac{1}{4}\)