Working with axiomatic probability. I am doing a course in discrete mathematics. One of the topics that our course touches upon is probability. Our syllabus doesn't require us to learn a lot of theory, that is to say that we use intuitive notions to solve most problems. As I was reading about probability on the internet I came across the axiomatic approach to probability which sounded very interesting. So I read quite a bit about it. What I am now trying to do is trying to solve homework problems using this approach to see if I truly understand what the axioms say and now I am stuck.

Problem: Consider a box which contains $4$ white , $5$ blue balls. What is the probability of drawing a white ball and a blue ball in some order if two balls are drawn given that each ball has an equal likelihood of being drawn?

I think I first have to understand what the experiment is.

Experiment: There are 4 white balls and 5 blue balls in a box. Two of these balls will be picked up randomly.

1. What is the sample space?
I think it should be:
$S={\{WB,BW,WW,BB\}}$
2.What is the event space?
It is the power set of $S$.
3. What is the event whose probability I wish to find?
Kolmogorov's system treats probability as a function whose argument is an event which is an element of the power set of the sample space. Here, the element that interests me is
$A=\{{WB,BW}\}$.
4. What is the value of $P(A)$?
$P(A)=P(\{WB,BW\})$
Since $\{WB,BW\}=\{WB\} \cup \{BW\}$ and because these two sets are disjoint, we conclude from the third axiom that $P(A)=P(\{WB\})+P(\{BW\})$.
5.What is the value of $P(\{WB\})$ ?
This is where I am stuck. I know from my intutition that the answer should be $\Large \frac{4}{9} \times \frac{5}{8}=\frac{5}{18}$. I do not know how to deduce this from the axioms. Any help would be appreciated.
 A: I preassume that the balls are drawn without replacement.
Let $A:=\{1,2,3,4,5,6,7,8,9\}$ and $\triangle:=\{\langle i,i\rangle\mid i\in A\}\subset A$
You can take $S=A^2-\triangle$ under the convention that $1,2,3,4$ correspond with white balls and the other numbers with blue balls.
Then $\wp(S)$ contains $9^2-9=72$ equiprobable singletons.
Observe that  $WB:=\{\langle i,j\rangle\in S\mid i\leq4\wedge j\geq5\}$ has $4\times5=20$ elements. 
So it is the union of $20$ singletons each having probability $\frac1{72}$.
This tells us that $\Pr(WB)=\frac{20}{72}=\frac5{18}$.
A: You need instead to define your sample space, $S$ as all different permutations of $WWWWBBBBB$ and then since each ball is equally to be chosen so is every permutation likely to be picked.(You would also need to prove this fact using your axioms but its very easy and trivial)
Now we care only for the first $2$ to be either $WB$ or $BW$ to be the first $2$ options  so our event that we need to find is the union of all the elements of $S$ that start with $WB$ or $BW$. 
Now,to side track a little bit we know that there are $9C5$ elements in the set $S$ and there are $2(7C4)$ elements in our event. And as we know $P(S) = 1 = P(\cup X_i)$ where $X_i$ denotes the $i$th permutation of $WWWWBBBBB$. And since $X_i = X_j$ $\forall i,j$ and since they are disjoint (no two are the same and they are single element sets)  then $(9C5)P(X_i) = 1 \implies P(X_i) = 1/126$.
Now the $P(A)$ where $A$ is the set of elements of $S$ such that  the elements start with $WB$ or $BW$ is equal to $2(7C4)P(X_i) = \frac{70}{126} = \frac{10}{18}$ which is eactly what you would expect.
