Probability marbles in a box with replacement

Louise Johnson

Junior Member
Joined
Jan 21, 2007
Messages
103
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
Joined
Jan 29, 2005
Messages
8,322
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
Joined
Jan 21, 2007
Messages
103
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
Joined
Jan 29, 2005
Messages
8,322
If we find P(GGR)+P(GRG)+P(RGG), that is like multiplying what you did by 3.
 

Louise Johnson

Junior Member
Joined
Jan 21, 2007
Messages
103
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
Joined
Jun 15, 2018
Messages
2
Correct answers are:
a) \frac{2}{{27}}[/tex]

b) \frac{2}{{3}}[/tex]
 

abinashnk

New member
Joined
Jun 15, 2018
Messages
2
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
 
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