Balls with duplicate colors grouped into groups of 5, probability of a group with $\ge 2$ colors

Question

This arose out of an online game, and both exact answers and approximations would be greatly appreciated.

$280$ balls are randomly put into $56$ bins such that all bins contain exactly $5$ balls. Among all $280$ balls, $2$ of them are colored red, $2$ of them green, $2$ of them blue (the rest can be considered uncolored, or any other color). I want to know the probability that, among the $56$ bins of balls, there exists $\ge 1$ bin which satisfies the following condition: the bin contains balls of at least two colors among the colors red, green, and blue.

For example, if we use "O" to denote a ball that is not colored as red, green, or blue. $\{R, B, O, O, O\}$ and $\{R,R,G,O,O\}$ are bins that satisfy the condition, while $\{R,R,O,O,O\}$ and $\{B,O,O,O,O\}$ do not.

My Attempt

I think an exact answer can be arrived (using multinomials), but I could not proceed beyond writing out the denominator. I decided to do a Poisson approximation, where $n = 56$ and $p$ is the probability that, when we randomly sample $5$ balls out of $280$, the $5$ balls satisfy the condition.

I calculated $p$ as follows:

$$
1 – \frac {
\binom{274}{5} + \binom{6}{1} \binom{274}{4} + 3 \binom{2}{2} \binom{274}{3}
} {\binom{280}{5}}
$$

And from then on I used $\lambda = n p$ and used $1 – e^{-\lambda}$ as the final probability that there exists $\ge 1$ bin which satisfies the condition.

• Is my $p$ correct, or did I count it wrong?
• Can I use Poisson approximation here? I know that Poisson can be used for weakly dependent events, but I am not sure this qualifies.

Edit: The Randomization Process

I realized that I probably should have stated the randomization process. The original game was randomized by randomly sampling 5 out of 280 into the first bin, 5 out of the remaining 275 into the second, 5 out of the remaining 270 into the third, and so on.

Answer 1

I approached this by drawing a tree diagram, starting with $R_1$ in any bin.

Adding $G_1$ has two possible outcomes: (Note that these are not $\frac{1}{56}$ and $\frac{55}{56}$)

• $Node_1$ has $G_1$ in same bin as $R_1$ with probability $\color{green}{P_1=\frac{4}{279}}$
• $Node_2$ has $G_1$ in new bin with probability $P_2=\frac{275}{279}$

The first case satisfies the condition of having at least two colors in one bin, so we stop there and add this to a list of successes. Continuing from $Node_2$, adding $B_1$ has three possible outcomes:

• $Node_3$ has $B_1$ with $R_1$, probability $\color{green}{P_3=\frac{4}{278}}$
• $Node_4$ has $B_1$ with $G_1$, probability $\color{green}{P_4=\frac{4}{278}}$
• $Node_5$ has $B_1$ in new bin, probability $P_5=\frac{270}{278}$

From $Node_5$, adding $R_2$ has four possible outcomes:

• $Node_6$ has $R_2$ with $R_1$, probability $P_6=\frac{4}{277}$
• $Node_7$ has $R_2$ with $G_1$, probability $\color{green}{P_7=\frac{4}{277}}$
• $Node_8$ has $R_2$ with $B_1$, probability $\color{green}{P_8=\frac{4}{277}}$
• $Node_9$ has $R_2$ in new bin, probability $P_9=\frac{265}{277}$

From $Node_6$, adding $G_2$ has four possible outcomes:

• $Node_{10}$ has $G_2$ with $R_1R_2$, probability $\color{green}{P_{10}=\frac{3}{276}}$
• $Node_{11}$ has $G_2$ with $G_1$, probability $P_{11}=\frac{4}{276}$
• $Node_{12}$ has $G_2$ with $B_1$, probability $\color{green}{P_{12}=\frac{4}{276}}$
• $Node_{13}$ has $G_2$ in new bin, probability $P_{13}=\frac{265}{276}$

From $Node_9$, adding $G_2$ has five possible outcomes:

• $Node_{14}$ has $G_2$ with $R_1$, probability $\color{green}{P_{14}=\frac{4}{276}}$
• $Node_{15}$ has $G_2$ with $R_2$, probability $\color{green}{P_{15}=\frac{4}{276}}$
• $Node_{16}$ has $G_2$ with $G_1$, probability $P_{16}=\frac{4}{276}$
• $Node_{17}$ has $G_2$ with $B_1$, probability $\color{green}{P_{17}=\frac{4}{276}}$
• $Node_{18}$ has $G_2$ in new bin, probability $P_{18}=\frac{260}{276}$

From $Node_{11}$, adding $B_2$ has four possible outcomes:

• $Node_{19}$ has $B_2$ with $R_1R_2$, probability $\color{green}{P_{19}=\frac{3}{275}}$
• $Node_{20}$ has $B_2$ with $G_1G_2$, probability $\color{green}{P_{20}=\frac{3}{275}}$
• $Node_{21}$ has $B_2$ with $B_1$, probability $P_{21}=\frac{4}{275}$
• $Node_{22}$ has $B_2$ in new bin, probability $P_{22}=\frac{265}{275}$

From $Node_{13}$, adding $B_2$ has five possible outcomes:

• $Node_{23}$ has $B_2$ with $R_1R_2$, probability $\color{green}{P_{23}=\frac{3}{275}}$
• $Node_{24}$ has $B_2$ with $G_1$, probability $\color{green}{P_{24}=\frac{4}{275}}$
• $Node_{25}$ has $B_2$ with $G_2$, probability $\color{green}{P_{25}=\frac{4}{275}}$
• $Node_{26}$ has $B_2$ with $B_1$, probability $P_{26}=\frac{4}{275}$
• $Node_{27}$ has $B_2$ in new bin, probability $P_{27}=\frac{260}{275}$

From $Node_{16}$, adding $B_2$ has five possible outcomes:

• $Node_{28}$ has $B_2$ with $R_1$, probability $\color{green}{P_{28}=\frac{4}{275}}$
• $Node_{29}$ has $B_2$ with $R_2$, probability $\color{green}{P_{29}=\frac{4}{275}}$
• $Node_{30}$ has $B_2$ with $G_1G_2$, probability $\color{green}{P_{30}=\frac{3}{275}}$
• $Node_{31}$ has $B_2$ with $B_1$, probability $P_{31}=\frac{4}{275}$
• $Node_{32}$ has $B_2$ in new bin, probability $P_{32}=\frac{260}{275}$

From $Node_{18}$, adding $B_2$ has six possible outcomes:

• $Node_{33}$ has $B_2$ with $R_1$, probability $\color{green}{P_{33}=\frac{4}{275}}$
• $Node_{34}$ has $B_2$ with $R_2$, probability $\color{green}{P_{34}=\frac{4}{275}}$
• $Node_{35}$ has $B_2$ with $G_1$, probability $\color{green}{P_{35}=\frac{4}{275}}$
• $Node_{36}$ has $B_2$ with $G_2$, probability $\color{green}{P_{36}=\frac{4}{275}}$
• $Node_{37}$ has $B_2$ with $B_1$, probability $P_{37}=\frac{4}{275}$
• $Node_{38}$ has $B_2$ in new bin, probability $P_{38}=\frac{255}{275}$

That completes the tree. Now add up the successes, multiplying through each parent node. $$\begin{array}{rl} P_1= & \frac{4}{279}\\ P_2(P_3+P_4)= & \frac{275}{279}\cdot\frac{8}{278}\\ P_2 P_5(P_7+P_8)= & \frac{275}{279}\cdot\frac{270}{278}\cdot\frac{8}{277}\\ P_2 P_5 P_6(P_{10}+P_{12})= & \frac{275}{279}\cdot\frac{270}{278}\cdot\frac{4}{277}\cdot\frac{7}{276}\\ P_2 P_5 P_9(P_{14}+P_{15}+P_{17})= & \frac{275}{279}\cdot\frac{270}{278}\cdot\frac{265}{277}\cdot\frac{12}{276}\\ P_2 P_5 P_6 P_{11}(P_{19}+P_{20})= & \frac{275}{279}\cdot\frac{270}{278}\cdot\frac{4}{277}\cdot\frac{4}{276}\cdot\frac{6}{275}\\ P_2 P_5 P_6 P_{13}(P_{23}+P_{24}+P_{25})= & \frac{275}{279}\cdot\frac{270}{278}\cdot\frac{4}{277}\cdot\frac{265}{276}\cdot\frac{11}{275}\\ P_2 P_5 P_9 P_{16}(P_{28}+P_{29}+P_{30})= & \frac{275}{279}\cdot\frac{270}{278}\cdot\frac{265}{277}\cdot\frac{4}{276}\cdot\frac{11}{275}\\ P_2 P_5 P_9 P_{18}(P_{33}+P_{34}+P_{35}+P_{36})= & \frac{275}{279}\cdot\frac{270}{278}\cdot\frac{265}{277}\cdot\frac{260}{276}\cdot\frac{16}{275} \end{array}$$ The sum is $$\frac{4440294}{27452639}\approx 16.17438\%$$