Urn with $4$ Different Colored Balls that is Drawn $4$ Times with Replacement There is an urn that contains $1$ Green ball, $1$ Red ball, $1$ White ball, and $1$ Yellow ball. 
The urn is drawn 4 times with replacement. So
I am trying to calculate: 
the probability of the the 4 draws yielding exactly 2 pairs of colors (e.g. GGRR)  
and 
the probability of the 4 draws yielding 1 single pair of color and 2 single / different colors (e.g. GGRW) 
For the 2 pairs, my thinking is such that
${4\choose2}$ represents the numbers of ways a paired color combo can be drawn / arranged

${4\choose1}$ is for the 1st color of the paired color combo

${3\choose1}$ is for the 2nd color of the paired color combo

${16\choose4}$ is for the total number of possible slots (_ _ _ _, each _ being a possible of 4 colors)
$$\frac{{4\choose2}{4\choose1}{3\choose1}}{16\choose4}$$
I am not sure if this is right, but I think I am at least somewhat on the right track 
For the 1 pair + 2 singles, in all honesty, I am bit lost about how to begin. I am not sure whether I should use the nCk formula or go by counting and then adjust for slot placement.

Thank you in advance
 A: Welcome to MSE.  This is a good first post.  It's clearly stated, and you've shown your work.  Unfortunately, there are some problems with the math.
The event space can be taken to be the set of $4$-element sequences from the letters G,R,W,Y.  There are $4^4=256$ of these, since each of the $4$ elements of the sequence has $4$ possible values.  Now for the first problem, there are $\binom42=6$ pairs of colors.  How many sequences can we make with them?  By the formula for permutations with repetitions, this is $\frac{4!}{2!2!}=6$, so the probability is $$\frac{36}{256}=\frac9{64}$$
Can you do the second problem now?
A: If you say there are $4^4=256$ equally likely ways of drawing the balls, then the number of ways of drawing them with  patterns of

*

*$4,0,0,0$ is $\frac{4!}{1!3!}\frac{4!}{4!0!0!0!}=4$

*$3,1,0,0$ is $\frac{4!}{1!1!2!}\frac{4!}{3!1!0!0!}=48$

*$2,2,0,0$ is $\frac{4!}{2!2!}\frac{4!}{2!2!0!0!}=36$

*$2,1,1,0$ is $\frac{4!}{1!2!1!}\frac{4!}{2!1!1!0!}=144$

*$1,1,1,1$ is $\frac{4!}{4!}\frac{4!}{1!1!1!1!}=24$
which add up to $256$
So the answers to your questions are $\dfrac{36}{256}=\dfrac{9}{64}$ and $\dfrac{144}{256}=\dfrac{9}{16}$
