= Solution
\b[There is no such value of $p$ in the specified interval.] Indeed, $\mathbb P(Y=1)=pe^{-p}>0$, and the <joint probability mass function> gives
$$
\mathbb P(X=1\mid Y=1)=1\ne p=\mathbb P(X=1).
$$
This contradicts <independence> for every $0<p<0.8$. Equivalently, $(1,2)$ has joint <probability> zero but the product of its marginal <probabilities> is $p\,e^{-p}p^2/2>0$.
At the suggested value $r=e^{-r}$, the <covariance> is zero, but the same <conditional probability> contradiction still applies. This is an explicit example of how <uncorrelated Bernoulli and Poisson variables need not be independent>. Zero <covariance> controls one mixed <moment>, whereas <independence> requires the entire <joint probability mass function> to factor.
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