#### Louise Johnson

##### Junior Member

- Joined
- Jan 21, 2007

- Messages
- 103

A box contains four green marbles and two red marbles. A marble is drawn and then replaced. This procedure is repeated three times.

a) Find the probability of drawing two green and one red marble, if the marble is replaced after each draw.

my answer: \(\displaystyle \L\\\frac{4}{6} \times \frac{4}{6} \times \frac{1}{6} = \frac{2}{{27}}\)

b) Find the probability of drawing two green and one red marble, if the marbles are

**not**replaced after each draw.

My answer: \(\displaystyle \L\\\frac{4}{6} \times \frac{3}{5} \times \frac{2}{6} = \frac{2}{{15}}\)

This question seemed alittle too easy so I am suspicious that I am missing something or haven't understood the problem. Let me know what you think.

Thank you

sincerly

Louise