Exercise: An ecologist wants to examine the deer population in two different areas, area 1 and area 2. He assumes the number of deer, X and Y, in areas 1 and 2, has Poisson distribution.
a.
The ecologist calculates that the expected number of deer is $2$ in area $1$ and $4$ in area $2$. Find $P(X=2)$ and $P(X>=3)$, and find an approximate expression for $P(X = Y)$. Specify the expectation you must use to calculate $P(x = y)$.
I have solved this exercise as written below:
Definition:
A random variable $X$ is said to have Poisson distribution with parameter $\lambda (>0)$ if $p(x)$ of $X$ is given by
\begin{align}\\ & p(x; \lambda) = \frac{e^{-\lambda}{\lambda ^x}}{x!} \end{align}
For $\lambda_1 = 2$:
\begin{align}\\ & P(X=2)=\frac{e^{-2}{2^2}}{2!}=\frac{2}{e^2}=0,27067\\ & P(X=1)=\frac{e^{-2}{2^1}}{1!}=\frac{2}{e^2}=0,27067\\ & P(X=0)=\frac{e^{-2}{2^0}}{0!}=\frac{1}{e^2}=0,13533\\ \end{align}
For $P(X>=3)$: \begin{align}\\ & P(X>=3) = 1-P(X<=2) \\ & P(X>=3) = 1-P(X=0)-P(X=1)-P(X=2) \\ & P(X>=3) = 1-0,27067-0,27067-0,13533 \\ & P(X>=3) = 0,32333 \end{align}
I assume this is correct, but please correct me if I got anything wrong.
b.
If a deer is in such an area, the ecologist assumes a probability of 0.4, that it will be observed. If there are actually 5 deer in the area, what is the probability that three of them will be observed? What is the expected number of deer observed?
I assume this is a Binominal distibution, because it satisfies the four axioms. There are two possible outcomes; observed or not observed. And $P(1)=0,4$. But I don't know how to continue. What am I doing wrong?
There are $\binom 53=10$ ways to choose the $3$ deer that will be observed. The probability that they will be observed is $.4^3$ and the probability that the other $2$ will not be seen is $.6^2$ Therefore, the probability is $$10\left(\frac25\right)^3\left(\frac35\right)^2$$