Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Assume that on average, only $1$ in $80$ calls made by a teleseller can he/she approaches a potential client. So

1.) what is the probability that a teleseller fails to approach any potential client in $1000$ calls made,and

2.) what is the least number of calls that a teleseller has to make in order to give a probability greater than 0.9 of approaching at least one potential client

I have tried to solve the first part with poisson but not sure whether Poisson model is suitable for this cases. For the second one, i have no idea how to get the number of calls

share|cite|improve this question
what is the probabilty that a given (correct) answer to this question is accepted? – Seyhmus Güngören Nov 18 '12 at 16:16
up vote 1 down vote accepted

Yes, a Poisson model is appropriate here-you have many low probability events. For 1) we have $\lambda=1000\cdot \frac 1{80}=12.5$ What is P(0)? For 2) you need to find $\lambda$ so that $P(0) \lt 0.1$, then convert that into the number of calls.

share|cite|improve this answer
It seems $\lambda$ is give in the question. Why you need to find $\lambda$ again in part 2? Also how to convert into the number of calls? – Mathematics Nov 18 '12 at 16:42
@Mathematics: $\lambda=np$, where $p$ is probability of success, and $n$ is the number of trials. In first problem, $np=1000/80$. For second, you get to pick $n$. – André Nicolas Nov 18 '12 at 16:49

Your Answer


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.