TITLE: Probability of drawing all 4 balls
QUESTION [11 upvotes]: A Greek urn contains a red, blue, yellow, and orange ball. A ball is drawn from the urn at random and then replaced. If one does this $4$ times, what is the probability that all $4$ colors were selected?
I approached this questions by doing $(1/4)^4$ because there's always a $1/4$ chance of selected a specific color ball if it's replaced. I also tried doing if not the correct ball was selected; so I did $(3/4)^4$ but that didn't work either. What am I doing wrong?

REPLY [2 votes]: The existing solutions provide the correct probability, but do not directly answer the question "What am I doing wrong?"
$(1/4)^4$ is the probability of a specific sequence of draws such as:
red, blue, yellow, orange
blue, yellow, orange, red
yellow, orange, blue, red
The event that "all 4 colors were selected" would occur if any of these sequences occurred. So we must count the number of such sequences (4! = 24) and add up their probabilities, which yields $\displaystyle\frac{4!}{4^4}$