4.3.1 problem 38

Let $I_j$ be the indicator random variable for $j$-th person drawing the slip of paper containing their name.

Let $X={\sum }_{j=1}^n{I}_j$ be the number of people who draw their name. Then, by linearity of expectation, $\text{E}\left(X\right)=\text{E}\left({\sum }_{j=1}^n{I}_j\right)={\sum }_{j=1}^n\text{E}\left(I_j\right)={\sum }_{j=1}^n\frac{1}{n}=1$.