Solution
= Solution
\b[The statement is true.] Use the <union bound> on the complementary events:
$$
P\left(\bigcap_{k=1}^n A_k\right)
=1-P\left(\bigcup_{k=1}^n A_k^c\right)
\geq1-\sum_{k=1}^nP(A_k^c)
=\sum_{k=1}^nP(A_k)-(n-1)>0.
$$
This proves the claim for arbitrary events, with no <independence> requirement.