APPROACH #1:

\(P(BGG) = \frac{C^1_4*C^2_8}{C^3_{12}}=\frac{4*28}{220}=\frac{28}{55}\).


APPROACH #2:

\(P(BGG) =\frac{3!}{2!}*\frac{4}{12}*\frac{8}{11}*\frac{7}{10}=\frac{28}{55}\). We need to multiply by 3!/2! because BGG scenario can occur in several ways: BGG, GBG, GGB (permutations of 3 letters BGG).

Answer: B.

Hope it's clear.
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1st Blue disk can be selected in 4C1 = 4 ways
1st Green disk can be selected in 8C1 = 8 ways
2nd Green disk (without replacement) can be selected in 7C1 = 7 ways

1 blue disk can be selected out of total 12 disks in 12C1 = 12 ways
1 green disk can be selected out of remaining 11 disks in 11C1 = 11 ways
2nd green disk can be selected out of remaining 10 disks in 10C1 = 10 ways
Probability = (4*8*7)/(12*11*10)

But, the order of ONE blue and TWO green could be arranged in 3!/2! ways.
So total probability = 3!/2! * (4*8*7)/(12*11*10)
= 28/55
Hence option B.

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