22. Let $S$ be the set of points whose coordinates $x, y$ and $z$ are integers that satisfy $0 \leq x \leq 2,0 \leq y \leq 3$ and $0 \leq z \leq 4$. Two distinct points are randomly chosen from $S$. Find the probability that the midpoint of the two chosen points also belongs to $S$ ?
設 $S$ 爲座標 $x 、 y 、 z$ 皆臬整數, 且滿足 $0 \leq x \leq 2 、 0 \leq y \leq 3$ 和 $0 \leq z \leq 4$ 的點集。現從 $S$ 中隨意抽出兩個不同的點，求該兩點的中點亦屬於 $S$ 的概率。
22. $\frac{23}{177}$
22. Note that $S$ contains $3 \times 4 \times 5=60$ elements. Divide the points of $S$ into 8 categories according to the parity of each coordinate. For instance, 'EEO' refers to the points of $S$ whose $x$ - and $y$ coordinates are even and whose $z$-coordinate is odd (and similarly for the other combinations of $\mathrm{E}$ and $\mathrm{O}$ ). There are $2 \times 2 \times 2=8$ points in the 'EEO' category. In a similar manner we can work out the size of the other categories:
\begin{tabular}{|c|c|c|c|c|c|c|c|c|}
\hline Category & EEE & EEO & EOE & EOO & OEE & OEO & OOE & OOO \\
\hline Size & 12 & 8 & 12 & 8 & 6 & 4 & 6 & 4 \\
\hline
\end{tabular}

Note further that two points form a favourable outcome if and only if they are from the same category. It follows that the required probability is
$$
\frac{C_{2}^{12}+C_{2}^{8}+C_{2}^{12}+C_{2}^{8}+C_{2}^{6}+C_{2}^{4}+C_{2}^{6}+C_{2}^{4}}{C_{2}^{60}}=\frac{23}{177} .
$$