# Probability marbles in a box with replacement

#### Louise Johnson

##### Junior Member
Question:
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

#### pka

##### Elite Member
The answers you have are for only one case: GGR.
But there are three cases: GGR, GRG, RGG.
The statement simply says "two greens and a red".
It does not say what order.

#### Louise Johnson

##### Junior Member
Ok I can see exactly what your saying I missed. Now to make up for those other combinations do I simply multiply the final answer by 3! or do I multiply each indivdual drawing of a marble by 3?
Thank you
Louise

#### pka

##### Elite Member
If we find P(GGR)+P(GRG)+P(RGG), that is like multiplying what you did by 3.

#### Louise Johnson

##### Junior Member
Please correct me if I am wrong, I think I have understood what you have said. My answers are then:

a) 2/9

b) 2/5

Thank you
Louise

#### abinashnk

##### New member
a) \frac{2}{{27}}[/tex]

b) \frac{2}{{3}}[/tex]

#### abinashnk

##### New member
Correct Answers (I can explain if you want!!)

a) 4/6 * 4/6 * 2/6 = 4/27

b) 4/6 * 3/5 * 2/4 * 3 = 3/5