Are there $a,b,c,d \in \mathbb N$ such that $\frac{a + b}{a + b + c + d} < \frac{a}{a + c} < \frac{b}{b + d}$? Consider the following $2 \times 2$ contingency table:
\begin{array}{c|cc|c}
 & C & \overline C & \Sigma \\ \hline
V & 4000 & 3500 & 7500 \\
\overline V & 2000 & 500 & 2500 \\ \hline
\Sigma & 6000 & 4000 & 10000
\end{array}
In data mining, we informally say that the association rule $C \to V$ is misleading. We could justify this by observing that:
$$
\Pr[V \mid C] = \frac{4000}{6000} = 0.\overline{6} < 0.75 = \frac{7500}{10000} = \Pr[V]
$$
Alternatively, we could justify this by observing that:
$$
\Pr[V \mid C] = \frac{4000}{6000} = 0.\overline{6} < 0.875 = \frac{3500}{4000} = \Pr[V \mid \overline C]
$$
I'm trying to figure out which justification is more compelling. When I change the numbers in the contingency table, both inequalities always seem to be simultaneously satisfied. I can't seem to make one inequality true and the other false.

More generally, consider the contingency table:
\begin{array}{c|cc|c}
 & C & \overline C & \Sigma \\ \hline
V & a & b & a+b \\
\overline V & c & d & c + d\\ \hline
\Sigma & a + c & b + d & a + b + c + d
\end{array}

Question: Do there exist $a,b,c,d \in \mathbb N$ such that either one of the following two inequalities are satisfied:
  \begin{align*}
\frac{a + b}{a + b + c + d} &< \frac{a}{a + c} < \frac{b}{b + d} \tag{1} \\
\frac{b}{b + d} &< \frac{a}{a + c} < \frac{a + b}{a + b + c + d} \tag{2}
\end{align*}


No. In general, for $a,b,c,d\in \Bbb N$ with $\frac{a}{b} < \frac{c}{d}$, $$\frac{a}{b} <\frac{a+c}{b+d}<\frac{c}{d}.$$
Proof: The left inequality is equivalent to $$a(b+d) < b(a+c) \iff ad < bc \iff \frac{a}{b}<\frac{c}{d}.$$ Likewise for the right inequality.
