4. As shown in Figure 1, there is a unit cube container $A B C D-A_{1} B_{1} C_{1} D_{1}$ with a volume of 1. There are holes at the midpoints of edges $A B$, $B B_{1}$, and diagonal $B C_{1}$, denoted as $E$, $F$, and $G$ respectively. If this container can be placed in any orientation, then the maximum volume of water it can hold is ( ).
(A) $\frac{1}{2}$ cubic units
(B) $\frac{7}{8}$ cubic units.
(C) $\frac{11}{12}$ cubic units
(D) $\frac{47}{48}$ cubic units
4. (C).

Using the reverse deduction method: If taking (U), it would be equivalent to removing the volume of the small triangular prism $E-B G$ $=\frac{1}{3} \times \frac{1}{4} \times \frac{1}{2} \times \frac{1}{2}=\frac{1}{48}$ (Figure 5). However, since it is a liquid, the liquid surface will form a quadrilateral $E F N M$ passing through $\triangle E F G$, at which point the container can only be
$$
1-V_{E I F-M N N}=1-\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times 1=\frac{7}{8} .
$$

If the liquid surface is adjusted to $\triangle E B_{1} C$ (Figure 6), then the volume is
$$
1-V_{E-B B_{1} C} C=1-\frac{1}{3} \times \frac{1}{2} \times \frac{1}{2} \times 1 \times 1=\frac{11}{12} \text {. }
$$

Therefore, the answer is (C).