5. Given points $A(1,1), B(1 / 2,0), C(3 / 2,0)$, the line passing through points $A, B$ and the line passing through points $A, C$ together with the line $y=a(0<a<1)$ enclose a planar region $G$. It is known that the probability of any point in the planar rectangular region $\{(x, y) \mid 0<x<2,0<y<1\}$ entering region $G$ is $\frac{1}{16}$, then $a=$ $\qquad$
Harm $\frac{1}{2}$.
Analysis The equation of line $AB$ is $y=2x-1$, and the equation of line $AC$ is $y=-2x+3$. The intersection points of the line $y=a$ with them are $D\left(\frac{1-a}{2}, a\right), E\left(\frac{3-a}{2}, a\right)$. The area of $G$ is equal to the area of triangle $ADE$, which is $\frac{(1-a)^{2}}{2}$. Therefore,
$$
\frac{(1-a)^{2}}{4}=\frac{1}{16}
$$

Solving it, we get: $a=\frac{1}{2}$.