Urn with three kinds of elements: $2n$ independent trials

Question

I guess this should be a quite simple problem, but I really don't know how to solve it. I am sorry if it might be too trivial.

An urn contains $c$ balls of three different kinds. There are $\alpha$ balls of kind $A>0$, $\beta>0$ balls of kind $B$, $\gamma>0$ balls of kind $G$, and $c=\alpha+\beta+\gamma$.

If we perform $2n$ independent trials (i.e. with replacement) from the urn, what is the probability to get $n$ balls of kind $A$ and $n$ balls of kind $B$?

I attempted to attack the problem by means of the concept of opposite event, but then I get completely lost.

Answer 1

Hint

What is the probability of drawing a particular ball on first try? Does this probability change?

So probability of each trial is fixed and equal to $P(A)=a/c$ and $P(B)=b/c$.

Now first think what are the total possible ways your $2n$ trials may look like. This is $3^{2n}$. Why?

Now construct how many will be favorable. This is $^{2n}C_n$

Solution might look like $$^{2n}C_n(p_a)^n(p_b)^n$$

Answer 2

Here you are dealing with trinomial distribution.

By every trial probability of drawing a ball of kind $A$ is $p_A:=\frac{\alpha}{c}$.

By every trial probability of drawing a ball of kind $B$ is $p_B:=\frac{\beta}{c}$.

By every trial probability of drawing a ball of kind $C$ is $p_C:=\frac{\gamma}{c}$.

Then in general by $m=r+s+t$ trials the probability of drawing $r$ of kind $A$, $s$ of kind $B$ and $t$ of kind $C$ equals:$$\binom{m}{r,s,t}p_A^rp_B^sp_C^t$$

Here $\binom{m}{r,s,t}$ is a notation for $\frac{m!}{r!s!t!}$.

Apply that for $m=2n$, $r=s=n$ and $t=0$.