 Solution

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 $0\le P(B)\le 1$.