= Solution
{c}
Each $X_i$ has the same <probability distribution> as $-X_i$, and <independence> makes $S_m$ and $-S_m$ identically distributed. Thus their two strict tail <probabilities> are equal:
$$
\mathbb P(S_m>m/2)=\mathbb P(S_m<-m/2)=\frac12\mathbb P(|S_m|>m/2).
$$
Apply <Chebyshev's inequality> using the <mean> and <variance> from part (i):
$$
\boxed{\mathbb P(S_m>m/2)\leq\frac12\frac{2m/3}{(m/2)^2}=\frac4{3m}.}
$$
The equal-tail argument is essential: applying <Chebyshev's inequality> directly to the absolute tail alone gives the weaker bound $8/(3m)$.
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