6 Let $S=\{(x, y)|-2 \leqslant y \leqslant| x \mid,-2 \leqslant x \leqslant 2\}$, then when $(x, y) \in$ $S$, and 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 Prompt: As shown in the figure, $S$ is the square $-2 \leqslant$ $x \leqslant 2, -2 \leqslant y \leqslant 2$. The area below the graph of $y=|x|$, $A_{s}=16-4=12$.

The equation $t^{2}+(|x|-1) t+|y|-2=0$ having one root greater than 1 and one root less than 1 is equivalent to $\left(t_{1}-\right.$ 1) $\left(t_{2}-1\right)<0$, which means $t_{1} t_{2}-\left(t_{1}+t_{2}\right)+1<0$, or $|x|+|y|<2$. The part of this within $S$ (i.e., the shaded area in the figure) has an area of $A_{p}=8-$ $2=6$, and the required probability is $P=\frac{A_{p}}{A_{S}}=\frac{1}{2}$.