683. The probabilities of three events $A_{1}, A_{2}, A_{3}$, forming a complete group, are given: $p_{1}=P\left(A_{1}\right)=0.22, p_{2}=$ $=P\left(A_{2}\right)=0.31, p_{3}=P\left(A_{3}\right)=0.47$. Conduct five trials, in each of which one of the three considered events occurs.
Solution. According to the rule of this paragraph, we need to simulate a discrete random variable $\boldsymbol{X}$ with the following distribution:

$$
\begin{array}{cccc}
X & 1 & 2 & 3 \\
p & 0.22 & 0.31 & 0.47
\end{array}
$$

According to rule $\S 1$, we divide the interval (0,1) into three sub-intervals: $\Delta_{1}-(0 ; 0.22), \Delta_{2}-(0.22 ; 0.43), \Delta_{3}-(0.43 ; 1)$.

We select five random numbers from the table in Appendix 9, for example, 0.61, 0.19, 0.69, 0.04, 0.46.

The random number $r_{1}=0.61$ belongs to the interval $\Delta_{3}$, so $X=3$ and, consequently, event $A_{3}$ occurs. Similarly, we find the other events. In the end, we obtain the desired sequence of events: $A_{3}, A_{1}, A_{3}, A_{1}, A_{3}$.