Probability-Bayes Thereom, what it means 
can case B and condition A be thought of as events?
considering the case $P(A \vert B) \times P(B)$ Can w be thought of as case b $\cap$ condition A and $w+y$ be thought of as case B $\cap$ condition A $+$ case B $\cap$ $A^c$, and then as all the events in B since B contains both those events in B and not in A and those events in B and in A?
So is the whole $\frac{w}{w+y}$ the proportion of those in case B $\cap$ condition   A to all the outcomes in event B? 
 A: Yes.   You have : $$\require{cancel}\begin{align}\mathsf P(A\cap B) ~&=~ \dfrac{\mathsf P(A\cap B)}{\mathsf P(B)}\cdot\dfrac{\mathsf P(B)}{1} \\[1ex] &=~ \dfrac{\mathsf P(A\cap B)}{\mathsf P(A\cap B)+\mathsf P(\overline A\cap B)}\cdot\dfrac{\mathsf P(A\cap B)+\mathsf P(\overline A\cap B)}{\mathsf P(A\cap B)+\mathsf P(\overline A\cap B)+\mathsf P(A\cap \overline B)+\mathsf P(\overline A\cap \overline B)} \\[1ex] &=~\dfrac{w ~/\cancel{(w+x+y+z)}}{(w+x) ~/\cancel{(w+x+y+z)}}\cdot\dfrac{(w+x) ~/\cancel{(w+x+y+z)}}{ (w+x+y+z) ~/\cancel{(w+x+y+z)}} \\[1ex] &=~ \dfrac{w}{w+x}\cdot\dfrac{w+x}{w+x+y+z} \\[1ex] &=~ \dfrac{w}{w+x+y+z}\end{align}$$
We thus define the conditional probability $\mathsf P(A\mid B)$ to equal the expresssion $\dfrac{\mathsf P(A\cap B)}{\mathsf P(B)}$ (as long as $\mathsf P(B)\neq 0$) .   And likewise define $\mathsf P(B\mid A) = \mathsf P(A\cap B)\div \mathsf P(A)$ (so long as $\mathsf P(A)\neq 0$).
So obtaining that: $\mathsf P(A\cap B)=\mathsf P(A\mid B)\cdot\mathsf P(B)=\mathsf P(A)\cdot \mathsf P(B\mid A)$.
