Let us consider the following probabilities:

• $P(X=0, Y=0) = a$
• $P(X=1, Y=1) = b$

From these, we can deduce:\begin{align}P(X=0) &= a + \frac{2}{9} = P(Y=0) \\P(X=1) &= \frac{2}{9} + b = P(Y=1)\end{align}

Now, if we assume that $X$ and $Y$ are independent variables, then the joint probabilities can be expressed as the product of their individual probabilities. Therefore, we have:

\begin{align}P(X=0, Y=0) &= P(X=0) \cdot P(Y=0) \\a &= (a + \frac{2}{9})^2\end{align}

Expanding and rearranging this equation, we obtain a quadratic equation:

\begin{align}a^2 + \frac{4}{81} - \frac{5}{9}a &= 0\end{align}

Solving this quadratic equation, we find that $a$ can be either $\frac{1}{9}$ or $\frac{4}{9}$.

Similarly, for $b$, we have:

\begin{align}P(X=1, Y=1) &= P(X=1) \cdot P(Y=1) \\b &= (\frac{2}{9} + b)^2\end{align}

This simplifies to the same quadratic equation as for $a$, yielding the solutions $b = \frac{1}{9}$ or $b = \frac{4}{9}$.

Considering all possible combinations of $a$ and $b$, we find the following two viable pairs that satisfy the independence requirement:

• $a = \frac{1}{9}, b = \frac{4}{9}$
• $a = \frac{4}{9}, b = \frac{1}{9}$

These pairs represent the solutions under the assumption of independence between $X$ and $Y$.