Poisson distribution Question - Expectation 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.
 A: 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,
$p:=P(N=0)=e^{-0.1}$, 
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 
$P(Y=n)=p^n(1-p)$
which is a geometric random variable, so that
$E(Y) = p/(1-p)$
A: 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
$$
\binom{50}{1} p^1 (1-p)^{50-1}.
$$
