Question

A bag contains 4 red balls, 2 green balls, 3 yellow balls, and 5 blue balls. Find each probability for randomly removing balls with
replacement.

Removing a yellow, a red, a green, and a blue ball =

Removing a blue, a green, a green, and a yellow ball =

Removing a red, a red, a yellow, and a yellow ball =

Removing a green, a yellow, a yellow, and a red ball =

Answer 1

Answer:

$\frac{15}{4802}$, $\frac{15}{9604}$, $\frac{9}{2401}$, $\frac{9}{4802}$

Step-by-step explanation:

The bag has a total of (4+2+3+5) = 14 balls. Set up the proportions:

Red: $\frac{4}{14}$

green: $\frac{2}{14}$

yellow: $\frac{3}{14}$

blue: $\frac{5}{14}$

Now solve!

Removing 1 yellow, 1 red, 1 green, and 1 blue = $\frac{3}{14} \cdot \frac{4}{14} \cdot \frac{2}{14} \cdot \frac{5}{14} =\frac{15}{4802}$

Removing 1 blue, 1 green, 1 green, and 1 yellow = $\frac{5}{14} \cdot \frac{2}{14} \cdot \frac{2}{14} \cdot \frac{3}{14} =\frac{15}{9604}$

Removing 1 red, 1 red, 1 yellow, and 1 yellow = $\frac{4}{14} \cdot \frac{4}{14} \cdot \frac{3}{14} \cdot \frac{3}{14} =\frac{9}{2401}$

Removing 1 green, 1 yellow, 1 yellow, and 1 red = $\frac{2}{14} \cdot \frac{3}{14} \cdot \frac{3}{14} \cdot \frac{4}{14} =\frac{9}{4802}$