= Solution
{c}
\b[No: <pairwise independence> does not imply that the three are <independent events>.] Take $C=\{Y\text{ is even}\}$. Then $\mathbb P(C)=1/2$, and $\mathbb P(A\cap C)=9/36=1/4$ by <independence> of the two rolls. Also $B\cap C$ is precisely the event that both faces are even, so $\mathbb P(B\cap C)=1/4$. Thus all three pairs of <events> are <independent>.
But $A\cap B\cap C$ also consists of the nine even-even outcomes, so
$$
\boxed{\mathbb P(A\cap B\cap C)=\frac14\ne\frac18
=\mathbb P(A)\mathbb P(B)\mathbb P(C).}
$$
This violates the triple-intersection requirement for <independent events>.
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