# Probability of Marbles

#### Gamest

##### New member
A blg contains 5 red marbles, 3 blue marbles and 2 green marbles. If six marbles are drawn without replacement from the bag, what is the probability that two marbles of each color are drawn? Express you answer as a common fraction.

I know that the number of ways to pick the marbles is around 10c6, but i don't know how to find the answer.

#### galactus

##### Super Moderator
Staff member
We are choosing 2 from 5, 2 from 3, and 2 from 2.

$$\displaystyle \frac{\binom{5}{2}\cdot\binom{3}{2}\cdot\binom{2}{2}}{\binom{10}{6}}$$

Or, we can do it this way:

$$\displaystyle \frac{6!}{2!2!2!}\underbrace{\left(\frac{5}{10}\right)\left(\frac{4}{9}\right)}_{\text{red}}\overbrace{\left(\frac{3}{8}\right)\left(\frac{2}{7}\right)}^{\text{blue}}\underbrace{\left(\frac{2}{6}\right)\left(\frac{1}{5}\right)}_{\text{green}}$$

We multiply by $$\displaystyle \frac{6!}{2!2!2!}=90$$ because that is how many ways the choices can be arranged.

They may come out like RGRBGB or RRGGBB, and so on.