Suppose $A$ and $B$ are two events with $P(B) \in (0,1).$ Let $P(A \mid B) < P(A).$ Can we say that $P(A \mid B^c) > P(A^c)$?
What I can observe is as follows $:$
$$P(A \cap B^c) = P(A) - P(A \cap B) > P(A) - P(A) P(B) = P(A) P(B^c).$$ So $P(A \mid B^c) > P(A).$
Thanks in advance.
On
Consider the Lebesgue probability space $((0, 1], \mathscr{B}((0, 1]), \lambda)$. Take $A = (1/4, 5/8], B = (1/2, 1]$, then
$$P(A | B) = \frac{P(AB)}{P(B)} = \frac{1/8}{1/2} = \frac{1}{4} < P(A) = \frac{3}{8}.$$
On the other hand,
$$P(A|B^c) = \frac{P(AB^c)}{P(B^c)} = \frac{1/4}{1/2} = \frac{1}{2} < P(A^c) = \frac{5}{8}.$$
Therefore the inequality is not necessarily true.
Let us flip two dices together. Let $A$ be the event that both the outcomes are odd and $B$ be the event that both the outcomes are even. Then $\Bbb P(A) = \Bbb P(B) = \frac {1} {12}$ and $\Bbb P(A \cap B) = 0 < \Bbb P (A) \Bbb P (B) = \frac {1} {144}.$ Now \begin{align*} \Bbb P (A \cap B^c) & = \Bbb P (A) = \frac {1} {12} < \frac {121} {144} = \Bbb P (A^c) \Bbb P (B^c). \end{align*}