Consider a game that consists of dealing out three hands of three cards each from a deck of nine cards. The deck contains the three Aces, three Kings, and three Queens. What is the probability someone is dealt three of a kind?
Attempt: Let D= event 3 of a kind are dealt to someone.
Then there are (9 choose 3 ways set the cards in each deck x 3 choose 1 ways to pick a deck)
(9 choose 1 * 2 choose 2 ways to pick the other two cards)
I'm confused now though. I don't understand the logic
Total number of hands is $\begin{pmatrix} 9 \\3\end{pmatrix}\begin{pmatrix} 6 \\3\end{pmatrix}$.
Number of hands where person 1 gets 3 of a kind is $3\begin{pmatrix} 6 \\3\end{pmatrix}$. Similarly for the other two players.
Number of hands where each person gets 3 of a kind is $3!$.
The probability is $$\frac{3(3\begin{pmatrix} 6 \\3\end{pmatrix}-3!)+3!}{\begin{pmatrix} 9 \\3\end{pmatrix}\begin{pmatrix} 6 \\3\end{pmatrix}}=\frac{9\begin{pmatrix} 6 \\3\end{pmatrix}-12}{\begin{pmatrix} 9 \\3\end{pmatrix}\begin{pmatrix} 6 \\3\end{pmatrix}}.$$
I'll use counting where....... $$P = \frac{\text{successful outcomes}}{\text{total outcomes}}$$
$$P = \frac{3\cdot \binom{6}{3}+2\cdot 3 (\binom{6}{3} - 2)}{\binom{9}{3}\cdot \binom{6}{3}} = \frac{60 + 108}{84\cdot 20} = \frac{1}{10}$$
