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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 $d=pq$. ThenAll four joint probabilities therefore factor, proving independence. HenceThis 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.

Here is an explicit joint distribution, for $0<p<0.8$: assign mass $(1+p)e^{-p}-p$ at $(0,0)$, mass $p(1-e^{-p})$ at $(1,0)$, mass $p{e}^{-p}$ at $(1,1)$, and mass $e^{-p}{p}^k/k!$ at $(0,k)$ for $k\ge 2$; all other masses are zero. These are nonnegative: the first is decreasing in $p$, and at $p=0.8$ it is still positive. Summing over rows and columns gives a Bernoulli distribution with mean $p$ and a Poisson distribution with mean $p$, so the joint mass sums to one.

Only $(1,1)$ contributes to the mixed first moment, giving $\mathbb{E}[XY]=p{e}^{-p}$ and the covariance above. The strictly decreasing function $e^{-p}-p$ has a unique zero in $(0,0.8)$, approximately $0.567143$. At that parameter the variables are uncorrelated. They are nevertheless dependent, because $\mathbb{P}(X=1\mid Y=1)=1$ whereas $\mathbb{P}(X=1)=p<1$. The Bernoulli independence from zero covariance result requires both variables to be Bernoulli and does not apply here.

Two square-integrable random variables $X,Y$ are uncorrelated random variables when $\mathrm{Cov}(X,Y)=0$. Independent random variables with finite second moments are uncorrelated random variables, but zero covariance does not generally imply independence.

For an explicit counterexample, let $X$ take $-1$, $0$, $1$ with equal probabilities, and let $Y=X^2$. Then $\mathbb{E}X=\mathbb{E}{X}^3=0$, so $\mathrm{Cov}(X,Y)=\mathbb{E}{X}^3-\mathbb{E}X\mathbb{E}{X}^2=0$. Nevertheless, $Y$ is completely determined by $X$: $\mathbb{P}(Y=0\mid X=0)=1$, while $\mathbb{P}(Y=0)=1/3$. Thus these are uncorrelated random variables with statistical dependence. In this example, opposite contributions from the odd function $x^3$ cancel; the even function $x^2$ still retains information about $|X|$. The last panel of the covariance diagram shows the relationship.

There are useful special cases. Uncorrelated jointly normal variables are independent, and the Bernoulli independence from zero covariance result supplies another setting where the converse holds. Their distribution assumptions do the extra work; zero covariance alone does not.