Example 1. A cube toy made of uniform material, with the numbers $1,2,3,4,5$, 6 marked on each face. If this toy is tossed once, what is the probability that the face up is an odd number? (Textbook P92 Exercise 6 Question 4)
Because the cube toy is uniform, the possibility of each face landing up is equal when it is tossed, with a total of six basic events $(\mathrm{n}=6)$. “The face that lands up shows an odd number” includes 3 basic events $(m=3)$, so $\mathrm{P}(\mathrm{A}) = \frac{3}{6} = \frac{1}{2}$.
2. Tabulation Method
List the basic events in a table to identify patterns.
In Example 1, if the cube toy is tossed twice, what is the probability that the sum of the numbers on the faces that land up is 7?

Let the first toss result in $\mathrm{a}$ points, and the second toss result in $\mathrm{b}$ points $(1 \leqslant \mathrm{a}, \mathrm{b} \leqslant 6)$, then the basic events can be seen as ordered pairs of natural numbers (a, b). All basic events can be listed in the table below:
\begin{tabular}{cccc}
$(1,1)$ & $(1,2)$ & $(1,3)$ & $(1,4)$ \\
$(1,5)$ & $(1,6)$ & $(2,1)$ & $(2,2)$ \\
$(2,3)$ & $(2,4)$ & $(2,5)$ & $(2,6)$ \\
$\ldots \ldots$ & & & \\
$(6,1)$ & $(6,2)$ & $(6,3)$ & $(6,4)$ \\
$(6,5)$ & $(6,6)$ & &
\end{tabular}

From the table, we can see that there are 36 equally likely possible outcomes, and “the sum of the numbers is 7” includes 6 basic events (marked in bold in the table), so the required probability
$$
\text { is } \mathrm{P}(\mathrm{A})=\frac{6}{36}=\frac{1}{6} \text {. }
$$