Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
add comment

2 Answers

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

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

share|improve this answer
add comment

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 $$ \binom{50}{1} p^1 (1-p)^{50-1}. $$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.