<s>[INST] Find the following probability A bowl contains 16 chips, of which 6 are red, 7 are white and 3 are blue. If four chips are taken at random and without replacement, find the probability that there is at least 1 chip of each colour.
Can someone please give me a hint?
Thank you!
 [/INST] 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
$$
P = \frac{819}{1820} = \frac{9}{20}
$$
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