= Solution
{c}
Summing the <joint probability mass function> over $y$ with the <geometric series> gives the <marginal probability mass function>
$$
p_X(x)=\frac{(1-q)q^x}{1-q^{10}},\qquad x=0,\ldots,9.
$$
Divide the <joint probability mass function> by this positive <marginal probability mass function> to obtain the <conditional distribution>:
$$
\boxed{\mathbb P(Y=y\mid X=x)=(1-q^{10})q^{10y},\qquad y\ge0,\quad0\le x\le9.}
$$
It is the <geometric distribution> counting failures before the first success, with success <probability> $1-q^{10}$. For the other nonnegative integers $x\ge10$, $\mathbb P(X=x)=0$; the elementary <conditional probability> given $X=x$ is undefined and supplies no additional case.
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