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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.

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    $\begingroup$ You have double-counted the case of choosing the same color in the first part of your equation. For example, if you choose red and blue you counted four reds, but if you choose red and green you also counted four reds once more. The easy way is just think there are three possibilities $(1,3)(2,2)(3,1)$ after you choose your two colors and just don't count the four-same-color case in the beginning. $\endgroup$
    – cr001
    Sep 16 '17 at 21:26
  • $\begingroup$ Ok, I will try again with this in mind and make an edit. $\endgroup$
    – Remy
    Sep 16 '17 at 21:35
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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

$$ \frac{\binom{4}{2}\left[ \binom{8}{4} - 2 \right]}{\binom{16}{4}} = \frac{102}{455}. $$

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    $\begingroup$ Nicely explained. $\endgroup$ Sep 16 '17 at 22:13
  • $\begingroup$ Thank you, your alternative solution was actually one of my prior solutions. I see why it is correct now. $\endgroup$
    – Remy
    Sep 16 '17 at 22:24
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Hint:

Here r-for red,b-for blue,g-for green,y-for yellow Now cases are $$\left( rbbb \right) \left( rrbb \right) \left( rrrb \right) \\ \left( ryyy \right) \left( rryy \right) \left( rrry \right) \\ \left( rggg \right) \left( rrgg \right) \left( rrrg \right) \\ \left( byyy \right) \left( bbyy \right) \left( bbby \right) \\ \left( yggg \right) \left( yygg \right) \left( yyyg \right) \\ \left( bggg \right) \left( bbgg \right) \left( bbbg \right) $$

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