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I'm still confused about how to apply the Poisson Distribution, could you help me to explain how to solve the following problem?

A company department takes on average 2 new employees per year. New employees are taken independently at any time of the year.

1) In a given year, what is the probability that the department employs at least one new employee?

(I was thinking of X ~Pois(2), while P(X≥ 1), but don't know how to continue and reach the result :-( )

2) What is the probability that there are less than 3 new employees?

(was thinking of X ~Pois(2), while P(X ≤ 3) = fx(1) + fx (2) +fx(3), but again don't know how to continue :-( )

3) What is the probability that there are no new employees in two consecutive years?

(no clue with this one!)

Thank you so much for your help!

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The Poisson model gives a reasonably good fit for this type of situation. Let $X$ be the number hired in a year. We will assume that $X$ has Poisson distribution with parameter $\lambda=2$. Then for any non-negative integer $k$, we have $$\Pr(X=k)=e^{-\lambda}\frac{\lambda^k}{k!}.\tag{1}$$

1) We have $$\Pr(X\ge 1)=1-\Pr(X=0)=1-e^{-\lambda} \frac{\lambda^0}{0!}=1-e^{-\lambda}=e^{-2}.$$

2) We have $X\lt 3$ precisely if $X=0$ or $X=1$ or $X=2$. Use Formula (1) to find $\Pr(X=k)$ for $k=0$, $1$, and $2$, and add up.

Note that the question asked for the probability of less than $3$. With discrete distributions, we have to distinguish carefully between $\lt$ and $\le$.

3) There are various approaches. Let $X$ be the number hired in the first year, and $Y$ the number hired in the second year. Both $X$ and $Y$ have Poisson distribution with parameter $2$. In the Poisson model, $X$ and $Y$ are independent, so $$\Pr(X=0\cap Y=0)=\Pr(X=0)\Pr(Y=0).$$

Another approach to 3) is that under the Poisson model, the number hired in $2$ years has Poisson distribution with parameter $2\lambda$, that is, $4$. So the probability none is hired in $2$ years is $e^{-4}$.

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