Show that $P(A\mid B) > P(A\mid B^{c}) \implies P(B\mid A) > P(B\mid A^{c})$ My initial thought is to start from $P(A\mid B)$ and $P(A\mid B^{c})$ and transform them to expressions involving $P(B\mid A)$ and $P(B\mid A^{c})$ respectively.
$P(A\mid B) = \dfrac{P(A)P(B\mid A)}{P(B)}$
$P(A\mid B^{c}) = \dfrac{P(A) - P(B) + P(A^{c})P(B\mid A^{c})}{P(B^{c})}$
The expressions get quite messy when I try to compare them, and I am wondering if there is a better a way.
 A: To begin with, notice that
\begin{align*}
P(A|B) - P(A|B^{c}) > 0 & \Longleftrightarrow \frac{P(A\cap B)}{P(B)} - \frac{P(A\cap B^{c})}{P(B^{c})} > 0\\\\
& \Longleftrightarrow \frac{P(B^{c})P(A\cap B) - P(B)P(A\cap B^{c})}{P(B)P(B^{c})} > 0\\\\
& \Longleftrightarrow P(A\cap B) - P(B)(P(A\cap B) + P(A\cap B^{c})) > 0\\\\
& \Longleftrightarrow P(A\cap B) - P(B)P(A) > 0
\end{align*}
Now we can manipulate the second relation in order to obtain
\begin{align*}
P(B|A) - P(B|A^{c}) & = \frac{P(A\cap B)}{P(A)} - \frac{P(A^{c}\cap B)}{P(A^{c})}\\\\
& = \frac{P(A^{c})P(A\cap B) - P(A)P(A^{c}\cap B)}{P(A)P(A^{c})}\\\\
& = \frac{P(A\cap B) - P(A)(P(A\cap B) + P(A^{c}\cap B))}{P(A)P(A^{c}))}\\\\
& = \frac{P(A\cap B) - P(A)P(B)}{P(A)P(A^{c})}
\end{align*}
which is positive due to the first result.
A: From the given information we can infer from properties of complements that
$$P\left (A^c|B \right ) < P\left (A^c|B^c \right)$$
Thus
$$P\left ( A| B \right ) > P\left( A |B^c \right ) > P \left (A^c |B^c\right )> P \left ( A^c |B\right ) $$
$$\frac{P\left( A \cap B \right )}{P(B)} > \frac{P\left( A \cap B^c \right )}{P(B^c)} >\frac{P\left( A^c \cap B^c \right )}{P(B^c)}> \frac{P\left( A^c \cap B \right )}{P(B)} $$
$$\frac{P(A) P \left (B |A\right )}{P(B)} > \frac{P(A) P \left (B^c |A\right )}{P(B^c)} >\frac{P(A) P \left (B^c |A^c\right )}{P(B^c)}> \frac{P(A^c) P \left (B |A^c\right )}{P(B)} $$
$$\frac{P(A) P \left (B |A\right )}{P(B)} > \frac{P(A^c) P \left (B |A^c\right )}{P(B)}$$
$$P(A) P \left (B |A\right ) > P(A^c) P \left (B |A^c\right )$$
We can also see from the Law of Total Probability that
$$P\left ( A\right ) = P\left (B \right)P\left (A|B\right ) + P\left (B^c \right)P\left (A|B^c\right )$$
$$P\left ( A^c\right ) = P\left (B \right)P\left (A^c|B\right ) + P\left (B^c \right)P\left (A^c|B^c\right )$$
Which means that
$$P(A) > P(A^c)$$
Therefore,
$$ P \left (B |A\right ) > P \left (B |A^c\right )$$
