Solution
= Solution
With $p=(1+\varepsilon)\log n/(n-1)$, the <union bound> gives
$$
\boxed{\mathbb P(L=0)\le n(1-p)^{n-1}
\le ne^{-p(n-1)}=n^{-\varepsilon}.}
$$
= Solution
With $p=(1+\varepsilon)\log n/(n-1)$, the <union bound> gives
$$
\boxed{\mathbb P(L=0)\le n(1-p)^{n-1}
\le ne^{-p(n-1)}=n^{-\varepsilon}.}
$$