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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

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what is the probabilty that a given (correct) answer to this question is accepted? –  Seyhmus Güngören Nov 18 '12 at 16:16
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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.

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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
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