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