How Can I Find the Probability of Drawing This Sample? A random sample of size $6$ is selected with replacement from an urn that contains $10$ red, $5$ white and $5$ blue marbles. What is the probability that the sample contains $2$ marbles of each color?
This is what I got so far: 
Pr[2 Red]= $ (\frac{10}{20})^2 $
Pr[2 Blue] = Pr[2 White] = $ (\frac{5}{20})^2 $
Pr[Event] = $ (\frac{10}{20})^2\cdot(\frac{5}{20})^4 \approx 0.001$
 A: As indicated in the comments, you have to take into account the number of sequences in which two red marbles, two white marbles, and two blue marbles are selected.  Since the sequence has length $6$, the number of ways two of the six positions can be filled with a red marble is $\binom{6}{2}$.  The number of ways two of the remaining four positions can be filled with a white marble is $\binom{4}{2}$.  The number of ways the remaining two positions can be filled with two blue marbles is $\binom{2}{2}$.  
As you determined, 
\begin{align*}
P(\text{two red}) & = \left(\frac{10}{20}\right)^2 = \left(\frac{1}{2}\right)^2\\
P(\text{two white}) & = \left(\frac{5}{20}\right)^2 = \left(\frac{1}{4}\right)^2\\
P(\text{two blue}) & = \left(\frac{5}{20}\right)^2 = \left(\frac{1}{4}\right)^2
\end{align*}
Hence, the probability that two red marbles, two white marbles, and two marbles are selected when six marbles are selected from ten marbles, five white marbles, and five blue marbles with replacement is 
$$\binom{6}{2}\binom{4}{2}\binom{2}{2}\left(\frac{1}{2}\right)^2\left(\frac{1}{4}\right)^2\left(\frac{1}{4}\right)^2$$
Note.  The number 
$$\binom{6}{2}\binom{4}{2}\binom{2}{2} = \frac{6!}{2!4!} \cdot \frac{4!}{2!2!} \cdot \frac{2!}{2!0!} = \frac{6!}{2!2!2!}$$
is the multinomial coefficient of the term $r^2b^2g^2$ in the multinomial expansion $(r + b + g)^6$.  
