Total balls - 12.
1st pick ( I need a blue) P(Blue) = No of blue balls/ Total = 5/12
At this point I have a blue ball for sure as my first pick.
2nd Pick : P(Green) = N(Green)/Total-1 = 4/11
Again, now I have a blue first and the green second.
3rd pick: P(yellow) = N(yellow)/Total - 2 = 3/10
So, P(BGY) in that order = 5/12 * 4/11 * 3/10 = 1/22.
So the order of the balls should be BGY. Now since there can be 6 possibilities of drawing 1 B, 1 G and 1 Y balls, so probability of B then G and then Y should be (probability of drawing 1B, 1G and 1Y), divided by 6.
Since we consider only a particular order here, we won't have 6 ways of getting BGY, we have only one way of getting that order, BGY can be arranged among themselves in 3! = 6 ways, but that would consider all arrangements you have mentioned. (BYG,YGB etc)
The difference is we already account for the order by doing this. If that was not the case, the first ball we pick up would be any one of the 3 colors, so probability should be 1 because it will certainly be a B,G or a Y. But we do 5/12 for the first because we only want a blue ball as our first pick.
Get it? I am not sure if I explain it well here.
