10. Product of Digits (7th grade, 3 points). Nезнайка (lit. "Don't-know") was solving the following problem: "Given a set of 20 random digits, find the probability that the product of these digits ends in 0." Nезнайка reasoned as follows.

If the set contains the digit 0, then the product of the digits will definitely end in 0. The probability of this is \(1 - \left(\frac{9}{10}\right)^{20}\). If the number does not contain 0 (the probability of this is \(\left(\frac{9}{10}\right)^{20}\)), then, for the product to end in 0, it must contain at least one even digit and the digit 5.

Among the nine non-zero digits, five are odd. The probability that all digits are odd is \(\left(\frac{5}{9}\right)^{20}\), so the probability that there is at least one even digit is \(1 - \left(\frac{5}{9}\right)^{20}\).

Once we have confirmed that the number does not contain 0 but does contain an even digit, we need to ensure that it contains the digit 5. The conditional probability that there is no 5, given that there is an even digit, is \(\left(\frac{8}{9}\right)^{19}\), since one position is occupied by an even digit, so a 5 cannot be in that position, leaving 19 positions, and a 5 is absent if each of these positions contains one of the eight digits (not 5 and not 0). Therefore, the conditional probability that there is a 5 is \(1 - \left(\frac{8}{9}\right)^{19}\). Consequently, the desired probability is

\[
\left(1 - \left(\frac{9}{10}\right)^{20}\right) + \left(\frac{9}{10}\right)^{20} \cdot \left(1 - \left(\frac{5}{9}\right)^{20}\right) \cdot \left(1 - \left(\frac{8}{9}\right)^{19}\right) \approx 0.987
\]

Знайка (lit. "Know-it-all") conducted an experiment: he generated 1000 random sequences of 20 digits each on a computer and found the proportion of sequences where the product of the digits ended in 0. The experiment yielded a result close to Nезнайка's answer. The discrepancy was very small and could be random if Nезнайка solved the problem correctly. Or it could be non-random.

Repeat Знайка's experiment several times. Can the experiment confirm or refute Nезнайка's correctness? If Nезнайка is indeed wrong, how should the problem be solved correctly?
Solution. There is an error in Nезнayka's reasoning: if there is an even digit in the set, it does not mean that it is the only one. If there are several even digits, then a five cannot stand on any of the positions occupied by these digits, so the conditional probability that there is a five depends on how many even digits are in the set.

Znayka's experiment is apparently correct. However, it only gives an approximate value of the probability. The correct reasoning seems to be as follows.

Let us introduce events $A_{i}$ “the i-th digit of the set is divisible by 2” and $B_{j}$ “the j-th digit is divisible by 5”, $\mathrm{P}\left(A_{i}\right)=0.5, \mathrm{P}\left(B_{j}\right)=0.2$. Events $A_{i}$ are pairwise independent, events $B_{j}$ are pairwise independent, and events $A_{i}$ and $B_{j}$ for $i \neq j$ are also independent. Moreover, events $A_{i}$ and $B_{i}$ are independent because the probability of their simultaneous occurrence is 0.1.

From this, it follows that the events $A_{1}, \ldots, A_{20}, B_{1}, \ldots, B_{20}$ are independent in combination. Indeed, if we have some set of these events, then events with different indices are independent since they relate to different digits, and events $A_{i}$ and $B_{i}$ with the same index, as we noted, are also independent, so the probability of the intersection of these events is equal to the product of their probabilities.

The event of interest to us, “the product ends in the digit 0,” consists in the occurrence of at least one of the events $A_{i}$ and at least one of the events $B_{i}$. The probability that none of the $A_{i}$ occur is $\left(1-\frac{1}{2}\right)^{20}=\left(\frac{1}{2}\right)^{20}$, and the probability of the opposite event “at least one of the events $A_{i}$ occurs” is $1-\left(\frac{1}{2}\right)^{20}$. Similarly, the probability that none of the $B_{i}$ occur is $\left(1-\frac{1}{5}\right)^{20}=\left(\frac{4}{5}\right)^{20}$, and the probability of the opposite event “at least one of the events $B_{i}$ occurs” is $1-\left(\frac{4}{5}\right)^{20}$. Since $A_{i}$ and $B_{i}$ are independent, the desired probability is

$$
\left(1-\left(\frac{1}{2}\right)^{20}\right) \cdot\left(1-\left(\frac{4}{5}\right)^{20}\right)=\left(1-0.5^{20}\right) \cdot\left(1-0.8^{20}\right) \approx 0.98847
$$

Answer: approximately $0.988$.