Question 191: A die is rolled three times, the first time it shows $a$, the second time it shows $b$, and the third time it shows $c$. Assuming the die shows numbers from 1 to 6 with equal probability,
(1) Find the probability that $a b + 2 c \geq a b c$;
(2) Find the probability that $a b + 2 c$ and $2 a b c$ are coprime.
Question 191, Solution: (1) Notice
$$
\mathrm{ab}+2 \mathrm{c} \geq \mathrm{abc} \Leftrightarrow \frac{1}{\mathrm{c}}+\frac{2}{\mathrm{ab}} \geq 1
$$

When $c=1$, there are $6^{2}=36$ sets of $(a, b, c)$ satisfying (1); when $c=2$, there are $4+2+1+1=8$ sets of $(a, b, c)$ satisfying (1); when $c=3$, there are $3+1+1=5$ sets of $(a, b, c)$ satisfying (1); when $c=4$, there are $2+1=3$ sets of $(a, b, c)$ satisfying (1); when $c=5$, there are $2+1=3$ sets of $(a, b, c)$ satisfying (1); when $c=6$, there are $2+1=3$ sets of $(a, b, c)$ satisfying (1).
Therefore, $\mathrm{P}(\mathrm{ab}+2 \mathrm{c} \geq \mathrm{abc})=\frac{36+8+5+3+3+3}{6^{3}}=\frac{29}{108}$
(2) Notice
$$
\begin{array}{l}
(a b+2 c, 2 a b c)=1 \\
\Leftrightarrow(a b+2 c, 2)=(a b+2 c, a)=(a b+2 c, b)=(a b+2 c, c)=1 \\
\Leftrightarrow\left\{\begin{array}{l}
a, b \in\{1,3,5\} \\
(a, c)=(b, c)=1
\end{array}\right.
\end{array}
$$

When $(a, b)=(1,1)$, there are 6 sets of $(a, b, c)$ satisfying (2); when $(a, b)=(3,3)$, there are 4 sets of $(a, b, c)$ satisfying (2); when $(a, b)=(5,5)$, there are 5 sets of $(a, b, c)$ satisfying (2); when $\{a, b\}=\{1,3\}$, there are $2 \times 4=8$ sets of $(a, b, c)$ satisfying (2); when $\{a, b\}=\{1,5\}$, there are $2 \times$ $5=10$ sets of $(a, b, c)$ satisfying (2); when $\{a, b\}=\{3,5\}$, there are $2 \times 3=6$ sets of $(a, b, c)$ satisfying (2).
Therefore, $P((a b+2 c, 2 a b c)=1)=\frac{6+4+5+8+10+6}{6^{3}}=\frac{13}{72}$.