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

`(15)/(4802)`,
`(15)/(9604)`,
`(9)/(2401)`,
`(9)/(4802)`

Explanation:

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

Red:
`(4)/(14)`

green:
`(2)/(14)`

yellow:
`(3)/(14)`

blue:
`(5)/(14)`

Now solve!

Removing 1 yellow, 1 red, 1 green, and 1 blue =
`(3)/(14) \cdot (4)/(14) \cdot (2)/(14) \cdot (5)/(14) =(15)/(4802)`

Removing 1 blue, 1 green, 1 green, and 1 yellow =
`(5)/(14) \cdot (2)/(14) \cdot (2)/(14) \cdot (3)/(14) =(15)/(9604)`

Removing 1 red, 1 red, 1 yellow, and 1 yellow =
`(4)/(14) \cdot (4)/(14) \cdot (3)/(14) \cdot (3)/(14) =(9)/(2401)`

Removing 1 green, 1 yellow, 1 yellow, and 1 red =
`(2)/(14) \cdot (3)/(14) \cdot (3)/(14) \cdot (4)/(14) =(9)/(4802)`