7. The expected value of the smallest number in a subset formed by randomly selecting $r(1 \leqslant r \leqslant n)$ elements from $\{1,2, \cdots, n\}$ is
Solution: The number of subsets of $\mathrm{M}$ containing $\mathrm{r}$ elements is $\boldsymbol{C}_{n}^{r}$, and the number of subsets of $\mathrm{M}$ containing $\mathrm{r}$ elements with the smallest number being the positive integer $\mathrm{k}$ is $\boldsymbol{C}_{n-\mathrm{k}}^{r-1}$, where $1 \leqslant \mathrm{k} \leqslant \mathrm{n}-\mathrm{r}+1$. Therefore, the expected value of the smallest number is
$$
\begin{array}{c}
\frac{1 \times C_{n-1}^{r-1}+2 \times C_{n-2}^{r-1}+3 \times C_{n-3}^{r-1}+\cdots+(n-r+1) \times C_{r-1}^{r-1}}{C_{n}^{r}} \\
=\frac{\left(C_{n-1}^{r-1}+C_{n-2}^{r-1}+\cdots+C_{r-1}^{r-1}\right)+\left(C_{n-2}^{r-1}+\cdots+C_{r-1}^{r-1}\right)+\cdots+C_{r-1}^{r-1}}{C_{n}^{r}} \\
=\frac{C_{n}^{r}+C_{n-1}^{r}+\cdots+C_{r-1}^{r-1}}{C_{n}^{r}}=\frac{C_{n+1}^{r+1}}{C_{n}^{r}}=\frac{n+1}{r+1} .
\end{array}
$$