Rich has 3 green, 2 red and 3 blue balls in a bag. He randomly picks 5 from the bag WITH replacement. What is the probability that of the 5 drawn balls, Rich has picked 1 red, 2 green, and 2 blue balls?
Choices.
A) (2/8)∗(3/8)^2∗(3/8)^2∗5!/(2!∗2!)  B) (2/8)∗(3/8)^2∗(3/8)^2∗5!  C) (2/8)∗(3/8)^2∗(3/8)2  D) (28)∗(38)∗(38)^2(28)∗(38)∗(38)^2  E) (38)2∗(38
CASE I.   Rich has 3 green, 2 red and 3 blue balls in a bag. He randomly picks 5 from the bag WITH replacement. What is the probability that of the 5 drawn balls, Rich has picked 1 red, 2 green, and 2 blue balls?   The probability would be: 2/8*3/8*3/8*3/8*3/8*5!/2!2!=14.8%   CORRECT (2/8)∗(3/8)^2∗(3/8)^2∗5!/(2!∗2!)NSWER:(2/8)∗(3/8)^2∗(3/8)^2∗5!/(2!∗2!)
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Rich has 3 green, 2 red and 3 blue balls in a bag. He randomly picks 5 from the bag WITH replacement. What is the probability that of the 5 drawn balls, Rich has picked 1 red, 2 green, and 2 blue balls?