Question

A probability model includes P(red) = 27 2 7 and P(blue) = 314 3 14 . Which of the following probabilities could complete the model? Select all that apply. A. P(green) = 27 2 7 , P(yellow) = 27 2 7 B. P(green) = 38 3 8 , P(yellow) = 18 1 8 C. P(green) = 14 1 4 , P(yellow) = 14 1 4 D. P(green) = 521 5 21 , P(yellow) = 1121 11 21 E. P(green) = 37 3 7 , P(yellow) = 114 1 14

Answer 1

Answer:

$(b)\ P(Green) = \frac{3}{8} ; P(Yellow) = \frac{1}{8}$

$(c)\ P(Green) = \frac{1}{4} ; P(Yellow) = \frac{1}{4}$

Step-by-step explanation:

Given

$P(Red) = \frac{2}{7}$

$P(Blue) = \frac{3}{14}$

Required

Which completes the model

Let the remaining probability be x.

Such that:

$P(Red) + P(Blue) + x = 1$

Make x the subject

$x = 1 - P(Red) - P(Blue)$

So, we have:

$x = 1 - \frac{2}{7} - \frac{3}{14}$

Solve

$x = \frac{14 - 4 - 3}{14}$

$x = \frac{7}{14}$

$x = \frac{1}{2}$

This mean that the remaining model must add up to 1/2

$(a)\ P(Green) = \frac{2}{7} ; P(Yellow) = \frac{2}{7}$

$P(Green) + P(Yellow)= \frac{2}{7} + \frac{2}{7}$

Take LCM

$P(Green) + P(Yellow)= \frac{2+2}{7}$

$P(Green) + P(Yellow)= \frac{4}{7}$

This is false because: $\frac{4}{7} \ne \frac{1}{2}$

$(b)\ P(Green) = \frac{3}{8} ; P(Yellow) = \frac{1}{8}$

$P(Green) + P(Yellow)= \frac{3}{8} + \frac{1}{8}$

Take LCM

$P(Green) + P(Yellow)= \frac{3+1}{8}$

$P(Green) + P(Yellow)= \frac{4}{8}$

$P(Green) + P(Yellow)= \frac{1}{2}$

This is true

$(c)\ P(Green) = \frac{1}{4} ; P(Yellow) = \frac{1}{4}$

$P(Green) + P(Yellow)= \frac{1}{4} + \frac{1}{4}$

Take LCM

$P(Green) + P(Yellow)= \frac{1+1}{4}$

$P(Green) + P(Yellow)= \frac{2}{4}$

$P(Green) + P(Yellow)= \frac{1}{2}$

This is true

Other options are also false

Answer 2

Answer:

B and C

Step-by-step explanation:

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