 Solution

The continuity of the cumulative distribution function gives the probability integral transform: $F(X_i)$ is uniform on $[0,1]$. Strict monotonicity or a probability density for $F$ is not required. Since $F$ is nondecreasing,By independence, $P(U\le u)=u^n$ for $0\le u\le 1$, so $U$ has probability density function $n{u}^{n-1}$. Average the conditional probability in part (b) using this distribution:Thus $\boxed{P(\tau =k)=n/[(k+n-1)(k+n)]}$. The expression telescopes to total mass one, so the waiting time is finite almost surely.