= Solution
{c}
Define $a_y=\min\{1,\lambda^y/7\}$. Given $Y=y$, the direct-return branch has <probability> $a_y$, while the fallback branch distributes its remaining mass uniformly over $0,\ldots,y$. The <law of total probability> therefore gives, for integer $x\geq0$,
$$
\boxed{\mathbb P(X=x)
=e^{-\lambda}\frac{\lambda^x}{x!}\min\!\left\{1,\frac{\lambda^x}{7}\right\}
+e^{-\lambda}\sum_{y=x}^{\infty}\frac{\lambda^y}{y!(y+1)}
\left(1-\min\!\left\{1,\frac{\lambda^y}{7}\right\}\right).}
$$
The summation includes $y=x$: a failed direct-return test can still return $x$ through the <discrete uniform distribution> draw. There is no rejection loop and no additional normalization. Summing the displayed <probability mass function> over $x$ gives $\sum_yp_y[a_y+(1-a_y)(y+1)/(y+1)]=1$. At $\lambda=1$, for example, $a_y=1/7$ at every $y$; it does not become one. This is an instance of the <accept-or-fallback mixture identity>.
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