Let $A$ be the event that first dice shows a $\var{number}$ $\Rightarrow P(A)=\frac{1}{6}$.
Let $B$ be the event that second dice shows a $\var{number}$ $\Rightarrow P(B)=\frac{1}{6}$.
$A$ and $B$ are independent events so $P(A\cap B) = P(A)\times P(B)$.
We want the probability $P(A \cup B)$ of either $A$ or $B$ showing $\var{number}$ and
\[\begin{eqnarray*}
P(A \cup B) &=& P(A)+P(B)-P(A \cap B)\\
&=& P(A)+P(B)-P(A)P(B)\\
&=&\frac{1}{6}+ \frac{1}{6}-\frac{1}{36}\\
&=& \frac{11}{36}
\end{eqnarray*}
\]