4.3.6 problem 43

(a)

This problem is a special case of problem 42 with $t=k$ and $n-1$ floors. Thus, the expected number of stops is $(n-1)-(n-1){\left(\frac{n-2}{n-1}\right)}^k$.

(b)

Let $I_j$ be the indicator variable for the $j$-th floor being selected for $2\le j\le n$. Then, the number of stops is $X={\sum }_{j=2}^n{I}_j$. Thus, $$\text{E}\left(X\right)=\text{E}\left(\sum _{j=2}^n{I}_j\right)=\sum _{j=2}^n\text{E}\left(I_j\right)=\sum _{j=2}^n(1-{(1-p_j)}^k)$$