An urn contains 5 red, 2 blue, and 9 green balls. Six balls are drawn.

Hint: the question you linked is looking at drawing balls without replacement. The way to approach the problem changes if you put each ball back into the urn after you draw it (i.e. with replacement), and the formula you used isn't correct for this situation.

Since there is replacement there is no conditional probability, this means that the answer is simply:

$$\frac{6!}{3!2!1!}p( \color{red}\bullet \cap \color{blue}\bullet \cap \color{blue} \bullet \cap\color{green}\bullet \cap\color{green}\bullet\cap\color{green}\bullet)=\frac{6!}{3!2!1!}p(\color{red}\bullet)p^2(\color{blue}\bullet )p^3(\color{green}\bullet \color{green})=60\left(\frac{5}{16}\right)\left(\frac{2}{16}\right)^2 \left(\frac{9}{16}\right)^3=\frac{54675}{1048576}\approx 5.21 \% $$

Where the factor $\frac{6!}{3!2!1!}$ indicates the number of permutations of the sequence of drawings.

:)