= Solution
{c}
For events $E_n$, the <Borel-Cantelli first lemma> states that $\sum_n\mathbb P(E_n)<\infty$ implies $\mathbb P(E_n\text{ infinitely often})=0$. Indeed,
$$
\mathbb P\left(\bigcup_{n\geq m}E_n\right)\leq\sum_{n\geq m}\mathbb P(E_n)\to0.
$$
These unions decrease to $\limsup E_n$, so continuity from above proves the assertion.
The <Borel-Cantelli second lemma> states that independent events with $\sum_n\mathbb P(E_n)=\infty$ satisfy $\mathbb P(E_n\text{ infinitely often})=1$. For every fixed $m$, independence and $1-x\leq e^{-x}$ give
$$
\mathbb P\left(\bigcap_{n=m}^N E_n^c\right)=\prod_{n=m}^N(1-\mathbb P(E_n))
\leq\exp\left(-\sum_{n=m}^N\mathbb P(E_n)\right)\to0.
$$
Continuity from above shows there is probability zero of no event after $m$. A countable union over $m$ gives probability zero of only finitely many occurrences. Independence is needed for this second conclusion, but not for the first.
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