Solution
= Solution
{c}
By <Chebyshev's inequality> and part (i),
$$
0\le\mathbb P\left(\frac{S_n}{n}>1+\epsilon\right)
\le\mathbb P\left(\left|\frac{S_n}{n}-1\right|\ge\epsilon\right)
\le\frac{\operatorname{Var}(S_n)}{n^2\epsilon^2}
=\frac{5n-2}{n^2\epsilon^2}\longrightarrow0.
$$
This actually proves the stronger statement $\boxed{S_n/n\xrightarrow{\mathbb P}1}$, namely <convergence in probability>, despite the <statistical dependence> between adjacent products.