= Solution
Write $q=P(B\mid A)$, $s=P(B\mid A^c)$ and $p=P(A)$. The <law of total probability> gives $P(B)=pq+(1-p)s$, a <convex combination> of the two fixed conditional values. Hence
$$
\boxed{\min(q,s)\leq P(B)\leq\max(q,s)}.
$$
These are the closed sharp bounds in the usual mixture-model interpretation, obtained by allowing $p$ to range from zero to one. The joint masses $pq,p(1-q),(1-p)s,(1-p)(1-s)$ realize every such mixture.
There is an endpoint qualification if both conditionals must literally be defined by ratios: then $0<p<1$. If $q\ne s$, the marginal lies strictly between them, so the stated endpoints are an <infimum> and <supremum> rather than achieved extrema. If $q=s$, every allowed mixture has exactly that common value. This is the second case of <sharp bounds from binary conditional probabilities>.
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