Your thought process seems to be the following:
- Pick the red chip
- Pick the white chip
- Pick the blue chip
- Pick one more chip
When you picked your one more chip however, the final chip will be an additional chip of an earlier color. Supposing that the chips were labeled with numbers as well as colors, the sequence of selections red1, white2, blue3, red4 would have resulted in the same collection of chips as red4, white2, blue3, red1.
Remember that to apply multiplication principle, we should have counted every outcome exactly once. Despite this, if instead we overcounted we can still salvage the approach so long as each outcome was counted exactly the same number of times each and we divide by the number of times each each counted as per the "shepherd's principle."
You can reason then that every outcome you will have counted exactly twice and so we should divide by two to account for that. Dividing by $\binom{16}{4}$, the number of ways of selecting four chips, will give the probability.
An alternate corrected approach which avoids division by symmetry arguments:
Recognize that to get at least one of each of the three colors represented in the four drawn chips exactly one of the following will occur:
- There will be two red chips, one blue, and one white
- There will be one red chip, two blue, and one white
- There will be one red chip, one blue, and two whites
Find the probabilities of each of these respective sub-cases and add.
$\dfrac{\binom{6}{2}\binom{7}{1}\binom{3}{1}+\binom{6}{1}\binom{7}{2}\binom{3}{1}+\binom{6}{1}\binom{7}{1}\binom{3}{2}}{\binom{16}{4}} = \frac{9}{20}$
Best Answer
I think barak manos is doubly counting some selections. Following Marconius's tip, we have
\begin{align} \binom{6}{2}(7)(3) + (6)\binom{7}{2}(3) + (6)(7)\binom{3}{2} & = (15)(7)(3) + (6)(21)(3) + (6)(7)(3) \\ & = 315 + 378 + 126 = 819 \end{align}
different ways to select chips to satisfy the condition. Note that there are
$$ \binom{6}{4} + \binom{7}{4} = 15 + 35 = 50 $$
different ways to select only red chips or only white chips, and that is the difference between this answer and barak's.
As in barak's answer, there are
$$ \binom{16}{4} = 1820 $$
different ways to select $4$ of $16$ chips, so the desired probability is
$$ P = \frac{819}{1820} = \frac{9}{20} $$