Probability of at least two balls of same color

Question

Let's assume we have $4$ sets of $8$ coloured balls: red, blue, black and white. If we draw $n$-times without replacing, where $2\leq n \leq 32$, what is the probability of getting at least 2 red balls?

My attempt is setting this up as a Laplace experiment. I will put different numbers on each ball, i.e. red balls get the numbers from $1$ to $8$, blue balls from $9$ to $16$ ... So the sample space $\Omega$ consists of $n$-tupels whose entries are pairwise different numbers. Each element $\omega \in \Omega$ has the same probabilty to get drawn, $p(\omega)=\frac{(32-n)!}{32!}$. Now I count all the possible elements in $\Omega$ which contain at least $2$ numbers in the range from $1$ to $8$: $${n \choose 2}\frac{8!}{(8-2)!}\frac{30!}{(30-n+2)!}$$ Then I receive the desired probability by: $$\frac{{n \choose 2}\frac{8!}{(8-2)!} \frac{30!}{(30-n+2)!}}{\frac{(32-n)!}{32!}}$$

Is this correct?

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Edit

This is my new approach:

To avoid double counting I separatly count all possible elements in $\Omega$ which contain exactly $2$, $3$, ...$n$ red balls:

$$\sum_{i=2}^8 {n \choose i}\frac{8!}{(8-i)!}\frac{24!}{(24-n+i)!}. $$

Where ${n \choose i}$ denotes the possibilites to select $i$-many spots among the $n$ draws and $\frac{8!}{(8-i)!}$ represents all ways to draw $i$-many red balls out of $8$ red balls in an ordered manner. Finally, $\frac{24!}{(24-n+i)!}$ counts all ways to fill up the remaining $(n-i)$ slots of the $n$-draws with balls that are not red. To receive the desired probability I divide by $\frac{32!}{(32-n)!}$ which yields: $$\frac{\sum_{i=2}^8 {n \choose i}\frac{8!}{(8-i)!}\frac{24!}{(24-n+i)!}}{\frac{32!}{(32-n)!}}. $$

Any comments are appreciated!

Answer 1

I would start with counting all possible (unordered) combinations with 0 and 1 red balls:

$$N_0=\binom{24}n;\quad N_1=\binom81\binom{24}{n-1}$$ and compute the probability as $$ 1-\frac {N_0+N_1}N, $$ where $$ N=\binom{32}n $$ is the overall number of combinations.

Answer 2

I couldn't follow Your logic, perhaps You could explain it better. Your probability equation gives $p>1$ and fails at $n=1$.

I think it's a lottery. We have $\binom{32}{n}$ possible draws, for $n\leq24$ there are $\binom{8}{0}\binom{24}{n}$ combinations with 0 red balls and for $n\leq25$ there are $\binom{8}{1}\times\binom{24}{n-1}$ combinations with 1 red ball.

$$p(r\geq2) = \begin{cases} 0 & \text{if $n=1$} \\ 1-\frac{\binom{24}{n}}{\binom{32}{n}}-\frac{8\times\binom{24}{n-1}}{\binom{32}{n}} & \text{if $2\leq n\leq 24$} \\ 1-\frac{8\times\binom{24}{24}}{\binom{32}{n}} & \text{if $n=25$} \\ 1 & \text{if $n>25$} \end{cases}$$