= Solution
\b[The statement is true.] Put $C_k=\bigcap_{r=1}^k A_r$. The assumed $P(C_{n-1})>0$ ensures that every conditioning event $C_{k-1}$, $2\leq k\leq n$, has positive probability. By the definition of <conditional probability>,
$$
P(A_k\mid C_{k-1})=\frac{P(C_k)}{P(C_{k-1})}.
$$
Multiplying makes the fractions telescope, giving
$$
\prod_{k=2}^nP(A_k\mid C_{k-1})=\frac{P(C_n)}{P(A_1)}
=P\left(\bigcap_{k=2}^n A_k\mid A_1\right).
$$
This proves the <chain rule for probabilities> here without any assumption of <independence>. The final numerator $P(C_n)$ may be zero; all denominators remain positive.
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