Probability of drawing all 4 balls 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?
 A: The probability of drawing $4$ different balls is the product of the probabilities of drawing a new ball on all $4$ draws.
The first draw yields a new ball, guaranteed:
$$P(\text{ball 1 new})=1$$
For the second draw, there are $3$ possible new balls and $4$ total balls, so:
$$P(\text{ball 2 new})=\frac34$$
For the third, there are $2$ possible new balls and $4$ total balls, so:
$$P(\text{ball 3 new})=\frac24=\frac12$$
For the fourth, there is one new ball and there are $4$ total balls, so:
$$P(\text{ball 4 new}) = \frac14$$

Thus, the answer is: $$\prod P = 1\cdot\frac34\cdot\frac12\cdot\frac14 = \frac3{32}=0.09375\text{ chance.}$$
A: The first ball can be any of the four with probability $\frac{4}{4}$
The second ball must be any of the other three with probability $\frac{3}{4}$
The third ball must be any of the other two with probability $\frac{2}{4}$
The fourth ball must be the ball that hasn't been selected yet with probability $\frac{1}{4}$
All together,
$$\frac{4}{4}\cdot \frac{3}{4}\cdot \frac{2}{4}\cdot\frac{1}{4}=0.09375$$
A: 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 $\displaystyle\frac{4!}{4^4}$
A: We could do this by counting the number of ways to draw four balls and the number of ways to draw four balls without getting any duplicates. 
There are $4!$ ways to not get a duplicate as every drawing can be thought of as an ordering and if we don't allow duplicates then we have a permutation. There are $4^4$ different possible drawings as replacement is allowed, this gives us $$\frac{4!}{4^4} = \frac{3}{32}$$
A: The first ball drawn can be any colour. So, the probability is $\frac{4}{4}.$
Since the first ball is replaced, there is a $\frac{1}{4}$ chance that the same ball will be drawn. The chance for a different ball to be drawn is $\frac {3}{4}$.
There is $\frac{2}{4}$ chance that the two drawn balls will duplicate, so the chance for a different ball to be drawn (for the third draw) is $\frac{2}{4}$.
Finally, there is a $\frac {1}{4}$ chance of the final different ball to be chosen as the other three are already drawn (if drawn) and will be duplicated.
$$\frac{4}{4} \times\frac{3}{4}\times\frac {2}{4}\times \frac {1}{4}=\frac{3}{32} =0.09375$$
Conclusion: There is a $\frac{3}{32} $  chance of you getting all four colours.
