Solve for a parameter and enforce probability bounds

$\begin{aligned}&\frac12\\ & =k+\frac16+k+\frac1{12}\\ &{}-\left(k+\frac16\right)\left(k+\frac1{12}\right).\end{aligned}$

Expanding gives $k^2-\frac74k+\frac{19}{72}=0$, or $72k^2-126k+19=0$.

$(12k-19)(6k-1)=0$, so $k=19/12$ or $1/6$.

From $0\le P(A)\le1$, $-1/6\le k\le5/6$. From $0\le P(B)\le1$, $-1/12\le k\le11/12$. Intersecting gives $-1/12\le k\le5/6$, which excludes $19/12$. Hence $k=1/6$.