16. Check Digit (8th grade. 3 points). The Simple-Dairy Telephone Company “Simple-Telecom” uses three-digit telephone numbers. The equipment is old, so during connection, errors in individual digits of the transmitted subscriber number are possible - each digit, independently of the others, has a probability of $p=0.02$ of being replaced by some other random digit. To reduce the probability of incorrect connection, the company uses the following rule: the first digit of the number is always equal to the remainder of the sum of the two other digits divided by 10. For example, the numbers 000 and 156 are possible, while the number 234 is not. If the check digit is incorrect during connection, an error message is displayed. Find the probability that, despite the measures taken, an incorrect connection will occur.
Solution. Numbers that satisfy the rule will be called correct. The task is to ensure that, despite the erroneous digits, the number turns out to be correct.

If two digits are known, then the third digit is uniquely determined. Therefore, we will not consider the first digit separately, but will assume that all digits are equal.

An erroneous number cannot be correct if it differs from the correct one by only one digit. Differences in two or three digits are possible. To find the probability that an erroneous number will turn out to be correct, note that the probabilities of two or three erroneous digits are respectively

$$
C_{3}^{2} p^{2}(1-p) \text { and } C_{3}^{3} p^{3} .
$$

Let $r_{2}$ be the probability that the number turned out to be correct given that it has exactly two erroneous digits. Similarly, $r_{3}$ is the probability that a number with three erroneous digits turned out to be correct. Then the total probability that an erroneous number is correct is

$$
C_{3}^{2} p^{2}(1-p) r_{2}+C_{3}^{3} p^{3} r_{3}
$$

Suppose there are two erroneous digits. No matter what the first one is, the second is uniquely determined: the event "the erroneous number is correct" is favorable only for one value of the second digit out of nine possible erroneous ones. Therefore, if two erroneous digits are fixed, the number will be correct with a probability of $r_{2}=\frac{1}{9}$.

Let's explain this with an example. Suppose in the number 376 errors occurred, for definiteness, in the 2nd and 3rd digits. Suppose the 2nd digit became the digit 5 by mistake. Then we get the number $35 \mathrm{~A}$, where A can take one of 9 values (all except 6) with equal chances. But the number will be correct only if $\mathrm{A}=8$, since only in this case will the sum of the last two digits end in the digit 3.

Suppose there are three erroneous digits. No matter what the first erroneous digit is, the event "the number will be correct" is favorable for 8 values of the second digit out of 9 equally possible erroneous values, and for the third digit - only 1 out of 9. Therefore, $r_{3}=\frac{8}{9} \cdot \frac{1}{9}$.

Let's explain this again with an example. Suppose in the number 376 the first digit became the digit 2 by mistake. Then we get the number $2 \mathrm{AB}$. The digit A is erroneous, so A can take any of 9 values (all digits except 7) with equal chances. But if A takes the value 6, the number will be correct only if $\mathrm{B}=6$, but this is impossible because $\mathrm{B}$ is also an erroneous digit and cannot equal 6. Therefore, the event "the number is correct" is favorable for 8 values of A out of 9 possible erroneous ones and only one value of B out of 9 possible. The probability of this is $\frac{8}{9} \cdot \frac{1}{9}$. Therefore,

$$
C_{3}^{2} p^{2}(1-p) r_{2}+C_{3}^{3} p^{3} r_{3}=3 \cdot 0.02^{2} \cdot 0.98 \cdot \frac{1}{9}+1 \cdot 0.02^{3} \cdot \frac{8}{81} \approx 0.000131
$$

Answer: approximately $0.000131$.