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Solving this by only applying principles of probability:
\(\\
\frac{3!}{2!} * \frac{8}{12} * \frac{7}{11} * \frac{4}{10} +\\
\frac{3!}{2! } * \frac{8}{12} * \frac{4}{11} * \frac{3}{10} +\\
\frac{3!}{3!} * \frac{8}{12} * \frac{7}{11} * \frac{6}{10}\)
\(= \frac{54}{55}\)

Logic here is that there are three cases which would allow at least one green bin to be selected - 1G bin, 2G bins and 3G bins (Remaining bins being selected as per remaining quantity). To quote Bunuel from another thread, we are multiplying by 3!2! each time since GRR scenario could occur in 3 ways: GRR, RGR, or RRG (the number of permutations of 3 letters GXX out of which 2 X's are identical), and so on for each case.
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