Is it possible to divide a square into four parts of arbitrary size with two lines? I am trying to visualize conditional probabilities. If you have two independent events $A$ and $B$, you can visualize the probabilities of all combinations $$A \cap B\qquad A \cap \neg B\qquad\neg A \cap B\qquad\neg A \cap \neg B$$ by just drawing the unit square and dividing it vertically with a line at $P(A)$ and horizontally with a line at $P(B)$. 
But if they're not independent it's... harder. But seems maybe possible with diagonal lines? Seems like perfect independence should be $90^{\circ}$ and perfect dependence should be $0^{\circ}$ parallel lines.
 A: Comment.
The usual representation is to use a vertical divider along with
horizontal lines at two appropriate heights. For example: $P(A)=.3, P(B|A)=.6,$ etc. could be represented as:

Unlike the usual Venn Diagrams, the idea is to represent probabilities by areas.
Yours is a clever idea, but I don't see how to draw an appropriate
diagonal line. I suspect it may be impossible in some cases, especially if the areas of the
left and right regions differ greatly. If you have a solution to this, I'd
be really happy to see it.
A: I consider this an elaboration of BruceET's answer. 
Given two lines in the plane, $S$ and $V$ let the areas of four regions they chop the unit square into be 
$\begin{pmatrix}A_{11}&A_{12}\\A_{21}&A_{22}\end{pmatrix}
=\begin{pmatrix}A_{11}(S,V)&A_{12}(S,V)\\A_{21}(S,V)&A_{22}(S,V)\end{pmatrix}$ 
(where $A_{11}(S,V)$ means the area above $S$ and to the left of $V$, etc).
Given four target numbers 
$\begin{pmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{pmatrix}$, whose sum is $1$,
where we assume all the $a_{ij}>0$ and that $a_{11}\lt a_{21}$,
we construct $S$ and $V$ so that all the $A_{ij}=a_{ij}$.
We pick $V$ to be the vertical line cut out by $x=a$ where $a=a_{11}+a_{21}$,
and exhibit a homotopy (indexed by $t\in[0,1]$) of lines $S(t)$,  so that the areas $A_{i1}(S,V) = a_{i1}$
for $i=1,2$ and so that $A_{21}(S,V)$ passes continuously from $1-a$ to $0$.
We specify the line $S$ by giving the coordinates of two points $P(t)$ and $Q(t)$ on $L(t)$, as functions of the homotopy index $t\in[0,1]$.   Let $Q(t)=(a,t)$.  
For $t\in[0,1-2a_{11}/a]$, let $P(t)=(a-2a_{11}/(1-t),1)$.
For $t\in[1-2a_{11}/a,0],$ let $P(t) = (0,2-2a_{11}/a - t).$ 
We see that for no value of $t$ is $P(t)=Q(t)$, so $S(t)$ is well-defined.
One checks that $A_{11}(S(t),V)=a_{11}$ for all $t$.
In words, the point $Q$ steadily rises along $V$. The point $Q$ starts on the top border of the square, moves left until it hits the upper left corner of the square, then descends along the left edge of the square. The slope of $S(t)$ starts negative and increases until it becomes positive. 
The line $S(t)$ sweeps over all the area in the square to the right of $V$. The area $A_{21}(S,V)$ to the right of $V$ and above $S(t)$ changes continuously in $t$, from $1-a$ down to $0$.   Along the way it hits $a_{21}$.
