= Solution
Independence requires each joint <probability> to equal the product of the relevant marginals. In particular it implies $d=(c+d)(b+d)$, equivalent to $ad=bc$.
Conversely let $p=c+d$ and $q=b+d$, and assume $ad=bc$. The <covariance> calculation gives $d=pq$. Then
$$
c=p-d=p(1-q),\qquad b=q-d=(1-p)q,\qquad a=1-b-c-d=(1-p)(1-q).
$$
All four joint <probabilities> therefore factor, proving <independence>. Hence
$$
\boxed{X,Y\text{ independent}\ \Longleftrightarrow\ ad=bc
\ \Longleftrightarrow\ X,Y\text{ uncorrelated}.}
$$
This is <Bernoulli independence from zero covariance>, including degenerate marginals. It is special to two-point variables and is not a general equivalence for arbitrary random variables.
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