6. Let $S=\{(x, y)|-2 \leqslant y \leqslant| x \mid,-2 \leqslant x \leqslant 2\}$.

Then when $(x, y) \in S$, and such that the quadratic equation
$$
t^{2}+(|x|-1) t+|y|-2=0
$$

has one root greater than $1$ and one root less than $1$, the probability is ( ).
(A) $\frac{1}{2}$
(B) $\frac{2}{3}$
(C) $\frac{3}{4}$
(D) 1
6. A.

As shown in Figure 2, S is a square
$$
\begin{array}{l}
-2 \leqslant x \leqslant 2, \\
-2 \leqslant y \leqslant 2
\end{array}
$$

The area below the graph of $y=|x|$, is
$$
A_{s}=16-4=12 \text {. }
$$

The equation
$$
t^{2}+(|x|-1) t+|y|-2=0
$$

has one root greater than 1 and one root less than 1
$$
\begin{array}{l}
\Leftrightarrow\left(t_{1}-1\right)\left(t_{2}-1\right)<0 \\
\Leftrightarrow t_{1} t_{2}-\left(t_{1}+t_{2}\right)+1<0 \\
\Leftrightarrow|x|+|y|<2 .
\end{array}
$$

The area of its part within $S$ is $A_{p}=8-2=6$. Therefore, the required probability is $P=\frac{A_{p}}{A_{s}}=\frac{1}{2}$.