Let us consider the following probabilities:

• $P(X=0,Y=0)=a$
• $P(X=1,Y=1)=b$

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

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:

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

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

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

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

Similarly, for $b$, we have:

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

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 $Y$.