= Solution
<Markov's inequality> states that for $Y\geq0$ and $t>0$, $\mathbb P(Y\geq t)\leq\mathbb E Y/t$. It follows by taking expectations in $Y\geq t\mathbf1_{\{Y\geq t\}}$.
If $X$ has finite <mean> $\mu$ and <variance> $\sigma^2$, apply Markov to $(X-\mu)^2$ to obtain <Chebyshev's inequality>:
$$
\boxed{\mathbb P(|X-\mu|\geq t)\leq\frac{\sigma^2}{t^2}\qquad(t>0).}
$$
For the <first moment method>, assume $\mu_n=\mathbb E X_n\to0$. Nonnegative <integer> values give $\{X_n>0\}=\{X_n\geq1\}$, so
$$
\boxed{\mathbb P(X_n>0)\leq\mathbb E X_n\longrightarrow0.}
$$
For a standard <second moment method>, suppose $\mu_n>0$ eventually, the <second moments> are finite, and $\mathbb E X_n^2/\mu_n^2\to1$. Equivalently, $\operatorname{Var}(X_n)/\mu_n^2\to0$. The zero <event> is contained in $\{|X_n-\mu_n|\geq\mu_n\}$, so
$$
\mathbb P(X_n=0)\leq\frac{\operatorname{Var}(X_n)}{\mu_n^2}\longrightarrow0.
$$
Hence $\boxed{\mathbb P(X_n>0)\to1}$. The second-moment condition controls the fluctuations relative to the <mean>; a large <mean> alone is insufficient.
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