My solution above is WRONG.
I’ve made train wrecks of these types of problems before.
I didn’t realize I’d wrecked the train this time until I read Catmg’s comment:
I think 9C3 is the number of ways to give out 3 cards to one person, not all 3.
That is true, and that means this question requires Hypergeometric distribution counts and NOT just Binomial distribution counts to correctly solve it.
The main difference between Hypergeometric distribution and Binomial distribution is that in Binomial distributions the samples sets are replaced before the next sample is drawn; in Hypergeometric distributions the samples are not replaced before the next sample is drawn.
This stands to reason: while any of the Binomial sets nCr(9,3) can exist, once a set is given to a person, the number and quality of sets remaining is greatly limited. For example: if person one receives three blue cards then person two cannot receive two yellows and a blue because there are no blue cards remaining to give.
Here is a solution using Hypergeometric distribution:
The number of possible dealt sets is \(N=\dbinom{9}{3} * \dbinom{6}{3} * \dbinom{3}{3} = 1680 \\\)
For person one, there are \(\dbinom{3}{1}\) ways to select one (1) of the three (3) red cards and then there are \(\dbinom{6}{2}\) ways to choose two more (non red) cards.
For person two, there are \(\dbinom{2}{1}\) ways to select one (1) of the two (2) remaining red cards and then there are \(\dbinom{4}{2}\) ways to choose two more (non red) cards.
For person three, select the remaining red card and the two remaining (non red) cards. There is one (1) way to do this.
Then
\(n = 3 \dbinom{6}{2} \cdot 2\dbinom{4}{2} \hspace {1em} \small | \text{ Where n = the number of hands with a red card.}\\ \Large \rho_{\small \text{(3 persons with one red card)}} \normalsize = \dfrac {n} {N} = 3! \cdot \dfrac{\dbinom{6}{2}\dbinom{4}{2}}{\dbinom{9}{3}\dbinom{6}{3}} = \dfrac{9}{56} \approx 16.07\%\\\)
GA
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