1587. How many times should a die be thrown so that the most probable number of times a two appears is 32?
Solution. In this case, $p=\frac{1}{6}, m_{0}=32$. It is required to find the number of independent trials $n$. The quantities $p, q=1-p$, $m_{0}$, and $n$ are related by inequality (1): $n p-q \leqslant m_{0} \leqslant n p+$ $+p$, from which

$$
n \cdot \frac{1}{6}-\frac{5}{6} \leqslant 32 ; n \cdot \frac{1}{6}+\frac{1}{6} \geqslant 32
$$

From the first inequality, $n \leqslant 197$, and from the second, $-n \geqslant 191$. Therefore, it is necessary to conduct from 191 to 197 independent trials.