= Solution
Summing the two nonzero entries with $X=1$ gives
$$
\mathbb P(X=1)=p(1-e^{-p})+pe^{-p}=p.
$$
The remaining <probability> is $\mathbb P(X=0)=1-p$, so \b[$X$ has the <Bernoulli distribution> with parameter $p$].
For the other marginal, the two entries at $Y=0$ add to $e^{-p}$, the entry at $Y=1$ is $pe^{-p}$, and for $k\geqslant2$ the only nonzero entry is already $e^{-p}p^k/k!$. Thus
$$
\boxed{\mathbb P(Y=k)=e^{-p}\frac{p^k}{k!},\qquad k=0,1,2,\ldots,}
$$
and \b[$Y$ has the <Poisson distribution> with parameter $p$]. This also checks normalization of the <joint probability mass function> by summing the marginal.
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