There are $\binom{6}{3}=20$ ways to choose three marbles from the bag.
We can consider the three colors separately:
- There are $\binom{2}{1}$ ways to choose one red marble,
$\binom{2}{1}$ ways to choose one blue marble, and
$\binom{2}{1}$ ways to choose one green marble.
So there are $\binom{2}{1} \cdot \binom{2}{1} \cdot \binom{2}{1} = 8$ ways to choose one marble of each color.
Therefore, the probability is $\frac{8}{20}=\boxed{\frac{2}{5}}$.
The answer is: \frac{2}{5}
Below is a math problem. Provide a detailed, step-by-step solution.

### Problem:
Three marbles are randomly selected, without replacement, from a bag containing two red, two blue and two green marbles. What is the probability that one marble of each color is selected? Express your answer as a common fraction.

### Answer:
