 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 $\boxed{P(B\mid A)>P(B)}$. Thus positive association of two events is symmetric.

For the complement of an event, $P(B^c)=1-b>0$, andTherefore $B^c$ 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.