= Solution
{c}
Summing over the last digit yields the <marginal probability mass function>
$$
p_Y(y)=(1-q)q^{10y}\sum_{x=0}^9q^x=(1-q^{10})q^{10y},\qquad y\ge0.
$$
Together with the other <marginal probability mass function>, this gives
$$
\boxed{p_{X,Y}(x,y)=p_X(x)p_Y(y).}
$$
The factorization holds on the whole support and with zeros elsewhere, proving that \b[$X$ and $Y$ are <independent>]. The <conditional distribution> in <4/c/ii> likewise does not depend on the permissible value $x$.
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