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A company takes on an average of $0.1$ new employees per week. Assume that the actual number of employees taken on in a given week has a Poisson distribution.

Let $Y$ denote the number of weeks that pass without a new employee being taken on. Calculate the expectation of $Y$. Calculate the probability that, over the course of a year ($50$ working weeks), there is just one week when more than one new employee is taken on.

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Assuming that each week is independent, we know that if $N$ is the number of employees taken on in a particular week, then $N$ has distribution:

$P(N=k) = e^{-0.1} \frac{0.1^k}{k!}$

So that,


which I'm calling $p$. Now, assuming we have an unlimited number of weeks, how can we not take on employees for exactly $Y$ weeks, no more no less.? Well we have to not take on employees for each week, a total of $Y$ weeks, but then take on at least one employee on the $Y+1$ week. The probability that we have 0 employees after $Y$ weeks is simply:

$P(\mbox{no employees after }Y\mbox{ weeks})=P(N=0)^Y=p^Y$

and the probability that we take on at least one employee on the $Y+1$ week is $1-P(N=0)=1-p$, so


which is a geometric random variable, so that

$E(Y) = p/(1-p)$

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Alex has answered part of the question.

The probability that more than one new employee is hired in a given week is $1$ minus the probability that only $0$ or $1$ new employees are hired. Thus it is $$ 1- \frac{0.1^0 e^{-0.1}}{0!} - \frac{0.1^1 e^{-0.1}}{1!} = 1 - e^{-0.1} - 0.1e^{-0.1}. $$ Call this number $p$. Then we're looking for the probability of $1$ success in $50$ trials with probability $p$ of success on each trial --- thus a binomial distribution. The probability is $$ \binom{50}{1} p^1 (1-p)^{50-1}. $$

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