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!
 A: 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
$$
P = \frac{819}{1820} = \frac{9}{20}
$$
A: Confirming Brian Tung's answer.
We have 16 chips; these are 6 red, 7 white, 3 blue


*

*There are $\dbinom{16}{4}$ ways to choose any 4 chips.


From these we need to subtract those which are not at-least one each kind.


*

*There are $\dbinom{10}{4}$ ways to choose 4 not-red chips.

*There are $\dbinom{9}{4}$ ways to choose 4 not-white chips.

*There are $\dbinom{13}{4}$ ways to choose 4 not-blue chips.
To avoid over counting we need to add back:


*

*There are $\dbinom{6}{4}$ ways to choose 4 only-red chips

*There are $\dbinom{7}{4}$ ways to choose 4 only-white chips

*There are $0$ ways to choose 4 only-blue chips (there's too few of them).
The principle of inclusion and exclusion (P.I.E.) then says:


*

*There are $\left[{\dbinom{16}{4}-\left[\dbinom{10}{4}+\dbinom{9}{4}+\dbinom{13}{4}\right]+\left[\dbinom{6}{4}+\dbinom{7}{4}\right]}\right]$ ways to choose 4 chips with at least one of each colour.


Divide that by $\dbinom{16}{4}$ to obtain the probability, $\tfrac 9{20}$ .

Alternatively (and more easily): since to have at least 1 of each colour, we need to choose 2 chips of one colour and 1 each of both others, then the probability is:
$$\dfrac{\dbinom{7}{2}\dbinom{6}{1}\dbinom{3}{1}+\dbinom{7}{1}\dbinom{6}{2}\dbinom{3}{1}+\dbinom{7}{1}\dbinom{6}{1}\dbinom{3}{2} }{ \dbinom{16}{4} } = \dfrac 9{20}$$
$\Box$
