5. Randomly draw three different numbers from $1,2, \cdots, 10$, the probability of the sample variance $s^{2} \leqslant 1$ is $\qquad$
5. $\frac{1}{15}$.

Notice that, for $x_{1}<x_{2}<x_{3}$, the sample variance is
$$
s^{2}=\frac{1}{3} \sum_{i=1}^{3}\left(x_{i}-\bar{x}\right)^{2} \leqslant 1
$$

if and only if $x_{1} 、 x_{2} 、 x_{3}$ are consecutive positive integers.
Thus, $P\left(s^{2} \leqslant 1\right)=\frac{8}{\mathrm{C}_{10}^{3}}=\frac{1}{15}$.