5. Given that a point $P$ is randomly chosen inside the tetrahedron $S-ABC$, the probability that $V_{\text {one-dimensional } P-ABC}<\frac{1}{2} V_{\text {three-dimensional } S-ABC}$ is ( ).
(A) $\frac{7}{8}$
(B) $\frac{3}{4}$
(C) $\frac{1}{2}$
(D) $\frac{1}{4}$
5. A.
$$
\Rightarrow h_{1}<\frac{1}{2} h_{2} \text {. }
$$

Thus, point $P$ is inside the frustum of a triangular pyramid $A_{1} B_{1} C_{1}-A B C$, as shown in Figure 6.
Notice that,
$$
S_{\triangle A_{1} B_{1} C_{1}}=\frac{1}{4} S_{\triangle A B C} \text {. }
$$

Therefore, the required probability is
$$
\begin{array}{l}
=1-\left(\frac{1}{2}\right)^{3}=\frac{7}{8} \text {. } \\
\end{array}
$$