= 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,
$$
U=F(Y_n)=\max_{1\leq i\leq n}F(X_i).
$$
By <independence>, $P(U\leq u)=u^n$ for $0\leq u\leq1$, so $U$ has <probability density function> $nu^{n-1}$. Average the conditional <probability> in part (b) using this distribution:
$$
P(\tau=k)=n\int_0^1u^{k+n-2}(1-u)\,du
=n\left(\frac1{k+n-1}-\frac1{k+n}\right).
$$
Thus \b[$\boxed{P(\tau=k)=n/[(k+n-1)(k+n)]}$]. The expression telescopes to total mass one, so the waiting time is finite almost surely.
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