A random variable $X = (X_1,X_2)$ has the following distribution (with $1<\theta<3$):
$$\begin{align}
\mathbb{P}(X_1 =x_1,X_2=x_2) = \begin{cases}\frac{1}{12}(12-7\theta+\theta^2) \quad (x_1,x_2)=(0,0), \\ \frac{\theta}{12}(4-\theta) \quad (x_1,x_2) = (0,1), \\ \frac{\theta}{12}(4-\theta)\quad (x_1,x_2) = (1,0), \\
\frac{\theta}{12}(\theta-1) \quad (x_1,x_2) = (1,1).
\end{cases}
\end{align}$$
Check whether $X_1 + X_2$ or $X_1X_2$ is sufficient for $\theta$.
All I know is that if $T(X)$ is a sufficient statistic of $\theta$, then we have that
$$\mathbb{P}(X_1 = x_1, X_2 = x_n | T(X) = t)$$
is independent of $\theta$.
I'm unsure how to do it here.
We have to compute the the conditional probabilities:
First,
$$P(X_1=i\cap X_2=j\mid X_1+X_2=k\ )=\begin{cases} &\text{ for }&k=0:\\ \\ 1&\text{ if }&i=0 \text { and }j=0\\ 0&\text{ if }&i=1 \text { and }j=0\\ 0&\text{ if }&i=0 \text { and }j=1\\ 0&\text{ if }&i=1 \text { and }j=1\\ \\ &\text{ for }&k=1:\\ \\ 0&\text{ if }&i=1 \text { and }j=1\\ 0&\text{ if }&i=0 \text { and }j=0\\ A(\theta)&\text{ if }&i=0 \text { and }j=1\\ B(\theta)&\text{ if }&i=1 \text { and }j=0\\ \\ &\text{ for }&k=2:\\ \\ 1&\text{ if }&i=1 \text { and }j=1\\ 0&\text{ if }&i=0 \text { and }j=0\\ 0&\text{ if }&i=0 \text { and }j=1\\ 0&\text{ if }&i=1 \text { and }j=0.\\ \end{cases}$$
All the probabilities above but $A(\theta)$ and $B(\theta)$ are obviously independent of $\theta$. So let's compute the missing probabilities: $$A(\theta)=P(X_1=0\cap X_2=1\mid X_1+X_2=1\ )=$$$$=\frac{P(X_1=0\cap X_2=1\cap X_1+X_2=1\ )}{P( X_1+X_2=1\ )}=\frac{P(X_1=0\cap X_2=1)}{P( X_1+X_2=1\ )}=$$ $$=\frac{P(X_1=0\cap X_2=1)}{P(X_1=0\cap X_2=1)+P(X_1=1\cap X_2=0)}=$$ $$=\frac{\frac{\theta}{12}(4-\theta)}{\frac{\theta}{12}(4-\theta)+\frac{\theta}{12}(4-\theta)}=\frac{1}{2}.$$
Notice that $$B(\theta)=A(\theta)=\frac12.$$ That is, so far, we've been talking about a sufficient statistic, $X_1+X_2$ on $\theta.$
Second,
$$P(X_1=i\cap X_2=j\mid X_1X_2=k\ )=\begin{cases} &\text{ for }&k=0:\\ \\ C(\theta)&\text{ if }&i=0 \text { and }j=0\\ D(\theta)&\text{ if }&i=1 \text { and }j=0\\ E(\theta)&\text{ if }&i=0 \text { and }j=1\\ 0&\text{ if }&i=1 \text { and }j=1\\ \\ &\text{ for }&k=1:\\ \\ 1&\text{ if }&i=1 \text { and }j=1\\ 0&\text{ if }&i=0 \text { and }j=0\\ 0&\text{ if }&i=0 \text { and }j=1\\ 0&\text{ if }&i=1 \text { and }j=0.\\ \end{cases}$$
Here $$C(\theta)=\frac{P(X_1=0\cap X_2=0)}{P(X_1=0\cap X_2=0)+P(X_1=0\cap X_2=1)+P(X_1=1\cap X_2=0)}=$$$$=\frac{\frac{1}{12}(12-7\theta+\theta^2)}{1-\frac{1}{12}(12-7\theta+\theta^2)}=\frac{(\theta-3)(\theta-4)}{7-\theta}$$ depending on $\theta$.
That is $X_1X_2$ is not a sufficient statistic on $\theta$.