62. In the city, there are 10,000 bicycles with various numbers from 1 to 10,000. What is the probability that the number of the first bicycle encountered will not contain the digit 8?
62. Since the first encountered cyclist can with equal probability turn out to be any of the 10000 cyclists in the city, the total number of equally probable outcomes of the experiment here is 10000. It remains to calculate the number of "favorable outcomes," i.e., the number of numbers that do not contain the digit 8. This number can be determined using the following method. We will conditionally prepend several zeros to all numbers containing fewer than four digits so that they become four-digit numbers (for example, instead of the number 26, we will write 0026), and replace the number 10000 with the number 0000. In this case, the number of numbers not containing the digit 8 will obviously remain the same. But now the numbers $0000, 0001, 0002, \ldots, 9999$ represent all possible four-digit numbers, where, in particular, a zero or several zeros can stand in the first place. If such a number does not contain the digit 8, then the first place can be occupied by any of the nine digits $0,1,2, \ldots, 7,9$; similarly, the second, third, and fourth places can be occupied by any of these nine digits. Combining nine possible values of the first digit with nine values of the second, third, and fourth digits, we will obviously get a total of $9^{4}$ four-digit numbers that do not contain the digit 8 (see the solution to problem 9).

Thus, out of 10000 equally probable outcomes (meeting any of the cyclists), $9^{4}$ outcomes will be favorable for us, from which it follows that the desired probability is $\frac{9^{4}}{10000}=(0.9)^{4}=0.6561$.