= Solution
Put $a=P(A)$, $b=P(B)$ and $j=P(A\cap B)$. Because $a,b>0$, the <conditional probability> inequality $P(A\mid B)>P(A)$ is equivalent to $j>ab$. But that same inequality gives $j/a>b$, or \b[$\boxed{P(B\mid A)>P(B)}$]. Thus <positive association of two events> is symmetric.
For the <complement of an event>, $P(B^c)=1-b>0$, and
$$
P(A\mid B^c)=\frac{a-j}{1-b}<\frac{a-ab}{1-b}=a.
$$
Therefore \b[$B^c$ strictly repels $A$]; equivalently $P(B^c\mid A)=1-j/a<1-b=P(B^c)$. Both directions are stated because the wording switches the grammatical roles of attraction and repulsion. The substantive conclusion is the same: the conditional probabilities involving the complement are strictly smaller than their unconditional counterparts.
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