Answer:
[tex](b)\ P(Green) = \frac{3}{8} ; P(Yellow) = \frac{1}{8}[/tex]
[tex](c)\ P(Green) = \frac{1}{4} ; P(Yellow) = \frac{1}{4}[/tex]
Step-by-step explanation:
Given
[tex]P(Red) = \frac{2}{7}[/tex]
[tex]P(Blue) = \frac{3}{14}[/tex]
Required
Which completes the model
Let the remaining probability be x.
Such that:
[tex]P(Red) + P(Blue) + x = 1[/tex]
Make x the subject
[tex]x = 1 - P(Red) - P(Blue)[/tex]
So, we have:
[tex]x = 1 - \frac{2}{7} - \frac{3}{14}[/tex]
Solve
[tex]x = \frac{14 - 4 - 3}{14}[/tex]
[tex]x = \frac{7}{14}[/tex]
[tex]x = \frac{1}{2}[/tex]
This mean that the remaining model must add up to 1/2
[tex](a)\ P(Green) = \frac{2}{7} ; P(Yellow) = \frac{2}{7}[/tex]
[tex]P(Green) + P(Yellow)= \frac{2}{7} + \frac{2}{7}[/tex]
Take LCM
[tex]P(Green) + P(Yellow)= \frac{2+2}{7}[/tex]
[tex]P(Green) + P(Yellow)= \frac{4}{7}[/tex]
This is false because: [tex]\frac{4}{7} \ne \frac{1}{2}[/tex]
[tex](b)\ P(Green) = \frac{3}{8} ; P(Yellow) = \frac{1}{8}[/tex]
[tex]P(Green) + P(Yellow)= \frac{3}{8} + \frac{1}{8}[/tex]
Take LCM
[tex]P(Green) + P(Yellow)= \frac{3+1}{8}[/tex]
[tex]P(Green) + P(Yellow)= \frac{4}{8}[/tex]
[tex]P(Green) + P(Yellow)= \frac{1}{2}[/tex]
This is true
[tex](c)\ P(Green) = \frac{1}{4} ; P(Yellow) = \frac{1}{4}[/tex]
[tex]P(Green) + P(Yellow)= \frac{1}{4} + \frac{1}{4}[/tex]
Take LCM
[tex]P(Green) + P(Yellow)= \frac{1+1}{4}[/tex]
[tex]P(Green) + P(Yellow)= \frac{2}{4}[/tex]
[tex]P(Green) + P(Yellow)= \frac{1}{2}[/tex]
This is true
Other options are also false
