Picking $4$ balls from $16$ balls. Chance that we get exactly $2$ different colors? Suppose $4$ balls will be randomly selected from a collection of $4$ red balls, $4$ blue balls, $4$ yellow balls, and $4$ green balls. What is the chance that exactly two of the colors will be present?
Attempted Solution:
Picking $2$ groups of colors out of the $4$ and then picking $4$ balls from the $8$ balls gives:
$4\choose2$$\frac{8\choose{4}}{16\choose4}$
But then there's the cases where all four balls balls selected from the $16$ are the same color giving:
$4\choose2$$\frac{8\choose{4}}{16\choose4}$ - $\frac{4\choose{1}}{16\choose4}$ = $.2286$.
Did I do this correctly? If so, and you have another method, I would be interested in seeing that as well.
 A: By selecting two colours and choosing 4 from 8, we can select 4 reds in three different ways:


*

*Select the colours red and blue and select 4 reds

*Select the colours red and yellow and select 4 reds

*Select the colours red and green and selected 4 reds
Therefore we are counting a selection of 4 reds three times. This is the same for the other 3 colours. Therefore the right correction is $3 \binom{4}{1}$. This gives a probability of
$$ \frac{\binom{4}{2}\binom{8}{4} - 3 \binom{4}{1}}{\binom{16}{4}} = \frac{102}{455}. $$
Alternatively, we can note that of the $\binom{8}{4}$ subsets we pick from the selection of two colours, 2 of them are inadmissible. Thus we can alternatively come to this answer as
$$ \frac{\binom{4}{2}\left[ \binom{8}{4} - 2 \right]}{\binom{16}{4}} = \frac{102}{455}. $$
A: Hint:
Here r-for red,b-for blue,g-for green,y-for yellow
Now cases are $$\left( rbbb \right) \left( rrbb \right) \left( rrrb \right) \\ \left( ryyy \right) \left( rryy \right) \left( rrry \right) \\ \left( rggg \right) \left( rrgg \right) \left( rrrg \right) \\ \left( byyy \right) \left( bbyy \right) \left( bbby \right) \\ \left( yggg \right) \left( yygg \right) \left( yyyg \right) \\ \left( bggg \right) \left( bbgg \right) \left( bbbg \right) $$
