Answer:
d) [tex]\frac{2}{3}[/tex]
Step-by-step explanation:
We have the following information:
total of marbles: 5 + 3 + 1 = 9
The probability of removing one marble from the bag and not being blue is:
[tex]P(NotBlue)= 1 - P(Blue)[/tex]
Where [tex]P(Blue)[/tex] is the probability of removing a blue marble, which is:
[tex]P(Blue)=\frac{BlueMarbles}{TotalMarbles} =\frac{3}{9} =\frac{1}{3}[/tex]
So, the probability of removing one marble that is not blue, is the total probability (1) minus the probability of removing a blue marble:
[tex]P(NotBlue)= 1 -\frac{1}{3}=\frac{2}{3}[/tex]
Another way to reach the same result is as follows:
[tex]P(NotBlue)=P(Red) + P(Green)[/tex]
Since it can't be a blue one, we add the chances of it coming out red or green.
[tex]P(Red)=\frac{RedMarbles}{TotalMarbles} =\frac{5}{9}[/tex]
[tex]P(Green)=\frac{GreenMarbles}{TotalMarbles} =\frac{1}{9}[/tex]
Thus:
[tex]P(NotBlue)=P(Red) + P(Green)=\frac{5}{9}+\frac{1}{9}=\frac{6}{9}=\frac{2}{3}[/tex]
Which is the same result we had found, so those are two ways of finding the answer to this problem.
