Probability of having at least one ball of each color An urn contains $6$ red balls, $2$ green balls and $9$ white balls. If a set of $4$ balls is randomly selected, but whenever a ball is selected its color is noted and then it is replaced in the urn before the next selection:
What is the probability that we have at least one ball of each color?
My attempt: Let
$E_{1}$= event of having $2$ red balls, $1$ green ball and $1$ white ball; 
$E_{2}$= event of having $1$ red ball, $2$ green balls and $1$ white ball; 
$E_{3}$= event of having $1$ red ball, $1$ green ball and $2$ white balls.
Then the proability $P$ (of having at least one ball of each color) is $=P(E_{1})+P(E_{2})+P(E_{3})$.
$$P(E_{1})=({6\over 17})^{2}{2\over 17}{9\over 17}$$
$$P(E_{2})={6\over 17}({2\over 17})^{2}{9\over 17}$$
$$P(E_{3})={6\over 17}{2\over 17}({9\over 17})^{2}$$
Then the probability $$P=({6\over 17})^{2}{2\over 17}{9\over 17}+{6\over 17}({2\over 17})^{2}{9\over 17}+{6\over 17}{2\over 17}({9\over 17})^{2}$$
The thing is that my friend told me that I need to multiply this number by $12$ but I don't see why I need to do this
Can you please tell me if this is the correct answer? I would appreciate your help :)
 A: Here the "pattern" $2$-$1$-$1$-$1$  is the only way our desired event can occur.  That makes your approach a good one. With larger numbers, the number of patterns may become large, and the above kind of analysis becomes unwieldy.  I suggest the following approach. 
Our desired probability is $1$ minus the probability that at least one colour is missing. We find that probability.
The probability that red is missing is $\left(\frac{11}{17}\right)^4$. We can find similar expressions for the probability green is missing, and for the probability that white is missing. 
Suppose we add together these three probabilities. Then we will have double-counted the situations in which two colours are missing. So we must subtract the sum of the "two colours missing" probabilities. For example, the probability that red and green are both missing is $\left(\frac{9}{17}\right)^4$.
The Inclusion/Exclusion procedure we have described is not much longer than yours for $4$ picks. And it remains essentially unchanged if we make, for example, $8$ picks: the exponents just change from $4$ to $8$. 
A: Yes, you should multiply by $\frac{4!}{2!}=12$, which is the number of ways to arrange $2$ red balls, $1$ green ball, and $1$ white ball in a line (if the red balls are indistinguishable) and the same goes for $E_2$ and $E_3$.
What you have computed in $P(E_1)$ is the probability of getting  $2$ red balls, $1$ green ball, and $1$ white ball in that order. For the other eleven orderings, the probability is the same.
A: You have to multiply by 12 because your sample space distinguishes between, say, the elements RRGW and WRRG. 
Thus the probability of getting a red, then a red, then a green, then a white IN THAT ORDER 
is 
($6\over7$)$^2$($2\over17$)($9\over17$).
But there's four unique ways to order four balls which are RRGW, each with the probability above, so you multiply by 4. And as you seem to understand, there's three ways to have four balls with at least one color of each, thus you multiply by 4 three times, or multiply by 12.
