Solution. We will use inequality (4.2.5). In this case, $n=10000, p=1 / 6, q=5 / 6$, so

$$
P\left(\left|\frac{m}{n}-p\right| \leq 0.01\right)=P\left(\left|\frac{m}{10000}-\frac{1}{6}\right|<0.01\right) \geq 1-\frac{\frac{1}{6} \cdot \frac{5}{6}}{10000 \cdot 0.01^{2}} \approx 0.86
$$
Example 1. Find the probability that the frequency of getting a six in 10000 independent tosses of a die deviates from the probability of getting a six by less than 0.01 in absolute value.