The probability of event $A$ occurring is $v=0.1$. We conduct $n=998$ trials. Is it more likely that $A$ occurs 100 times, or 99 times?
If the probability of a certain event occurring is $v$, then the probability that it will occur exactly $k$ times out of $n$ trials, specifically on the prescribed $n_{1^{-}}, n_{2}-, \ldots, n_{k}$-th trials, and not occur in the other $n-k$ trials, is clearly $v^{k}(1-v)^{n-k}$. If we do not specify the $k$ positions in the $n$ trials, i.e., any $k$-element combination of the first $n$ integers is acceptable, then the probability that the event will occur in some way $k$ times out of $n$ trials, and not occur $n-k$ times, is:

$$
V_{n / k}=\binom{n}{k} v^{k}(1-v)^{n-k}
$$

In our problem, we need to compare $V_{998 / 99}$ and $V_{998 / 100-a t}$, where $v=0.1$. It will be useful to determine the ratio $\frac{V_{n / k}}{V_{n / k+1}}$.

$$
\begin{gathered}
\frac{V_{n / k}}{V_{n / k+1}}=\frac{\binom{n}{k} v^{k}(1-v)^{n-k}}{\binom{n}{k+1} v^{k+1}(1-v)^{n-k-1}}= \\
=\frac{n(n+1 \ldots(n-k+1))}{1 \cdot 2 \cdot \ldots \cdot k} \cdot \frac{1 \cdot 2 \cdot \ldots \cdot k(k+1)(1-v)}{n(n-1) \ldots(n-k+1)(n-k) v}= \\
=\frac{(k+1)(1-v)}{(n-k) v}
\end{gathered}
$$

thus

$$
\frac{V_{998 / 99}}{V_{998 / 100}}=\frac{100 \cdot 0.9}{899 \cdot 0.1}=\frac{900}{899}>1
$$

and therefore

$$
V_{998 / 99}>V_{998 / 100}
$$

Remark: The result may seem surprising at first glance, because applying the law of large numbers superficially and incorrectly (as many solvers did), it appears that in 998 trials ( $998 \cdot 0.1=) 99.8$ is closer to 100 than to 99.

Let's see for what value of $k$ will $V_{n / k}>V_{n / k+1}$.

From the inequality

$$
\frac{V_{n / k}}{V_{n / k+1}}=\frac{(k+1)(1-v)}{(n-k) v}>1
$$

it follows that

$$
k>(n+1) v-1
$$

in which case $V_{n / k}>V_{n / k+1}$, or in the present case

$$
k>(998+1) 0.1-1=98.9
$$

Thus, $k=99$ is the smallest number for which $V_{n / k}>V_{n / k+1}$.