Consider an urn containing $4$ red balls, $4$ blue balls, $4$ yellow balls, and $4$ green balls. If 8 balls are randomly drawn from these 16 balls, what is the probability that it will contain at least one ball of each of the four colors?
Attempted Solution:
I think I did this right, but I just wanted to confirm.
P(at least one of each ball)
= $1$ – P(not one of each)
= $1$ – $\frac{12\choose8}{16\choose8}$ = $.9615$
Edit:
Actually, I think I have to choose one group of 4 to not get any selected, i.e. $4\choose1$ and then take
$1$ – $4$$\frac{12\choose8}{16\choose8}$ = $.846$
Best Answer
We can use inclusion-exclusion principle: Let $A_1,A_2,...,A_4$ denote events that there will be a red, blue, green, yellow ball in our selection respectively.
Then, by symmetry:$$P(A_1^c\cup...\cup A_4^c) = 4P(A_1^c)-6P(A_1^c\cap A_2^c) + 4P(A_1^c\cap A_2^c \cap A_3^c) - P(A_1^c\cap A_2^c \cap A_3^c \cap A_4^c) $$
Noting that $$P(A_1^c) = \frac{\binom{12}{8}}{\binom{16}{8}} \quad \quad P(A_1^c\cap A_2^c) = \frac{\binom{8}{8}}{\binom{16}{8}}$$ and the remaining terms are zeroes.
Thus $$P(A_1 \cap...\cap A_4) = 1-P(A_1^c\cup...\cup A_4^c) = \color{red}{\frac{1816}{2145}}=0.846620...$$
which is different from your proposed answer: $$1-4\frac{\binom{12}{8}}{\binom{16}{8}} = \frac{11}{13} = 0.846154...$$