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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. HenceThese 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.

Put $p=P(A)$ and $q=P(B\mid A)$. Decompose the marginal as $P(B)=P(A\cap B)+P(A^c\cap B)=pq+t$, where $0\le t\le 1-p$. It follows thatBoth bounds are attainable: choose none or all of the mass outside $A$ to belong to $B$. More explicitly, the four joint cell masses can be $pq,p(1-q),t,1-p-t$, all nonnegative. This is the first case of sharp bounds from binary conditional probabilities.

For the usual conditional-probability ratio, the given $q$ requires $p>0$. If $p=1$, the two bounds coincide at $q$. If $p=0$, the conditional is undefined as a ratio, although its formal joint-model bounds reduce to $0\le P(B)\le 1$.