Poisson Distribution?? finding the probability of randomly distributed trees. Assume that the aggressive invasive tree known as European Buckthorn is randomly distributed in a degraded forest preserve with **λ =40** trees/seedlings per **100m^2**. If **two 100 m^2** plots are randomly chosen then what is the probability that one of the plots has at **least 30** buckthorn trees/seedlings while the other plot has **30 or less** such trees? I guess it's Poisson distribution. However, how should it be solved when question asks to find one area with _at least 30_ trees while the other area is _30 or less_? Thank you!

We assume that the tree count in each plot is independent of the other. Let $N_1, N_2 \sim {\rm Poisson}(\lambda = 40)$ be IID Poisson variables indicating the random number of such trees in each plot. Then the desired probability is $\Pr[N_1 \ge 30 \cap N_2 \le 30] + \Pr[N_1 \le 30 \cap N_2 \ge 30].$ But because $N_1, N_2$ are IID, this probability is simply $$\begin{align*} 2 \Pr[N_1 \le 30] \Pr[N_2 \ge 30] &= 2 (\Pr[N_1 \le 30])(1 - (\Pr[N_2 \le 30] - \Pr[N_2 = 30])) \\\ &= 2p\left(1 - p + e^{-40} \frac{40^{30}}{30!}\right), \end{align*}$$ where $$p = \Pr[N_1 \le 30] = e^{-40} \sum_{i=1}^{30} \frac{40^i}{i!}.$$ This gives a probability of approximately $0.118054$.