Ball and urn problem Let's say there are three types of balls, labeled $A$, $B$, and $C$, in an urn filled with infinite balls. The probability of drawing $A$ is $0.1586$, the probability of drawing $B$ is $0.81859$, and the probability of drawing $C$ is $0.02275$.
If you draw $6$ balls, what's the probability that you drew $3 A$s, $2 B$s, and $1 C$?
I know what the answer is but I don't know how to set up the problem appropriately. It's not just a simple counting problem because the balls have weighted probabilities... Any push in the right direction is appreciated!
 A: The appropriate probability distribution for this question is the multinomial distribution, which is a generalization of the binomial distribution.  That is to say, let $\boldsymbol X = (A,B,C)$ be a vector-valued random variable that counts the number of balls drawn of each type. Then the parameters are $n = 6$, $p_1 = 0.1586$, $p_2 = 0.81859$, and $p_3 = 1 - p_1 - p_2 = 0.02275$.  Then $$\Pr[\boldsymbol X = (3,2,1)] = \frac{6!}{3!2!1!} p_1^3 p_2^2 p_3^1.$$
A: The result of a draw can be represented as a $6$-letter word over the alphabet $\{A,B,C\}$.
Let us first count the number of words over this alphabet that have $3$ $A$'s, $2$ $B$'s, and $1$ $C$.
The location of the $A$'s can be chosen in $\binom{6}{3}$ ways, and for each way the location of the $B$'s can be chosen in $\binom{3}{2}$ ways, for a total of $\binom{6}{3}\binom{3}{2}$.
Any particular such word has probability $p_A^3p_B^2p_C^1$, where $p_A$, $p_B$, and $p_C$ are the probabilities of drawing a ball of type $A$, of type $B$, of type $C$. For given the very large size of the urn, we can assume that the results of successive draws are independent. 
Thus the required probability is 
$$\binom{6}{3}\binom{3}{2}p_A^3p_B^2p_C^1.$$
A: Any sequence of the sort $ABACAB$ has the same probability to occur. So it comes to finding how many distinct words there are having $3$ times an $A$, $2$ times a $B$ and $1$ time a $C$. 
You could start with $6$ open spots and then placing the $A$'s. That gives $\binom{6}{3}$ possibilities. 
Then place the $B$'s. That gives $\binom{3}{2}$ and finally place $C$ on the single spot that is left. 
There are $\binom{6}{3}\times\binom{3}{2}=60$ possibilities so the corresponding probability is $60\times p_A^3p_B^2p_C$ 
