3. A fair cube die, with faces numbered $1, 2, 3, 4, 5, 6$, is thrown twice. Let the number on the first throw be $a$, and the number on the second throw be $b$. Then, the probability that the system of equations
$$
\left\{\begin{array}{l}
a x+b y=3, \\
x+2 y=2
\end{array}\right.
$$

has only positive solutions for $x$ and $y$ is ( ).
(A) $\frac{1}{12}$
(B) $\frac{2}{9}$
(C) $\frac{5}{18}$
(D) $\frac{13}{36}$
3. D.

When $2 a-b=0$, the system of equations has no solution.
When $2 a-b \neq 0$, the solution to the system of equations is
$$
(x, y)=\left(\frac{6-2 b}{2 a-b}, \frac{2 a-3}{2 a-b}\right) \text {. }
$$

From the given information, we have
$$
\begin{array}{l}
\left\{\begin{array}{l}
\frac{6-2 b}{2 a-b}>0, \\
\frac{2 a-3}{2 a-b}>0
\end{array}\right. \\
\Rightarrow\left\{\begin{array} { l } 
{ a = 2 , 3 , 4 , 5 , 6 } \\
{ b = 1 , 2 ; }
\end{array} \text { or } \left\{\begin{array}{l}
a=1, \\
b=4,5,6,
\end{array}\right.\right.
\end{array}
$$

There are $5 \times 2+3=13$ cases;
Since there are $6 \times 6$ $=36$ basic events when rolling the dice twice, the required probability is $\frac{13}{36}$.