202. Indicate the number of steps sufficient to assert with an error probability not exceeding 0.001 that the reduced speed of the chip is less than 0.01.

Let us recall now that each movement of the chip is conditioned by the result of tossing a coin. If, in $n$ tosses of the coin, heads appear $l$ times and tails appear $n-l$ times, the chip will make $l$ steps to the right and $n-l$ steps to the left and will end up at the point

$$
l-(n-l)=2 l-n
$$

The reduced speed of the chip over $n$ steps will be expressed by the absolute value of the ratio

$$
\frac{2 l-n}{n}=2 \frac{l}{n}-1
$$

The fraction $\frac{l}{n}$ characterizes the frequency of heads appearing.

Suppose a certain allowable error probability is given. As we know, for large values of $n$, it can be asserted with practical certainty that the reduced speed is close to zero. From expression (8), it is clear that if the reduced speed is small, then $2 \frac{l}{n}$ is approximately equal to 1 and, consequently, the frequency $\frac{l}{n}$ is close to $\frac{1}{2}$. Thus:

If a coin is tossed a large number of times, it is practically certain that the frequency of heads appearing will be close to $\frac{1}{2}$.

Roughly speaking, it is practically certain that heads will appear in about half of the cases. A more precise formulation states:

Choose an arbitrary allowable error probability $\varepsilon$ and specify any arbitrarily small number $\alpha$. If the number of coin tosses exceeds

$$
N=\frac{1}{\alpha^{2} \sqrt[3]{\varepsilon^{2}}}
$$

then with an error probability less than $\varepsilon$, it can be asserted that the frequency of heads appearing differs from $\frac{1}{2}$ by less than $\alpha$.

The proof of this precise formulation easily follows from statement b) on page 151: if $n>\frac{1}{\alpha^{2} \sqrt[3]{\varepsilon^{2}}}$, then

$$
\frac{1}{\sqrt{n}}<\alpha \cdot \sqrt[3]{e} \text { and } \frac{\frac{2}{\sqrt[3]{\varepsilon}}}{\sqrt{n}}<2 \alpha
$$

Thus, with an error probability less than $\varepsilon$, the reduced speed of the chip is less in absolute value than $2 \alpha$.

But the reduced speed in our case is the absolute value of $\frac{2 l-n}{n}=2 \frac{l}{n}-1$. Therefore, with an error probability less than $\varepsilon$, it can be asserted that $2 \frac{l}{n}$ differs from 1 by less than $2 \alpha$ or, in other words, $\frac{l}{n}$ differs from $\frac{1}{2}$ by less than $\alpha$.
202. With a probability of error less than 0.001, it can be asserted that the given speed of the chip after $n$ steps will be less than $\frac{2}{\sqrt[3]{0.001} \sqrt{n}}$.

Let's take such an $n$ so that

$$
\frac{2}{\sqrt[3]{0.001} \sqrt{n}} \leqslant 0.01
$$

from which $n \geqslant 4 \cdot 10^{6}$.

Thus, with a probability of error less than 0.001, it can be asserted that the given speed after $4 \cdot 10^{6}$ steps will be less than

$$
\frac{2}{\sqrt[3]{0.001} \sqrt{4 \cdot 10^{6}}}=0.01
$$