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If Phone calls are received to a switch board at a rate of four calls per hour. If the operator leaves his workstation for half an hour what is the probability he will miss a phone call?

I'm also confused to figure whether its related to Poisson distribution or Exponential distribution

**Update:**Sorry I'm still confused which distribution to use. My lecturer said exponential distribution can be used in these kind of scenarios. What is the most suitable type of distribution to be used in this kind of scenario and why?

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2 Answers 2

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In the OP, the question is raised about whether to use the Poisson or the exponential. In this question, you could use either.

Poisson: The number $X$ of phone calls in a half hour has Poisson distribution with parameter $4/2$. Thus $$\Pr(X\ge 1)=1-\Pr(X=0)=1-e^{-2}.$$

Exponential: Because the number of calls per hour has Poisson distribution with parameter $4$, the waiting time $W$ between calls has exponential distribution with mean $1/4$. So $W$ has density function $4e^{-4w}$ for $w\gt 0$.

It follows that $\Pr(W\le w)=1-e^{-4w}$. It follows that at least one call is missed (the waiting time is less than $1/2$) with probability $1-e^{-2}$.

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Let $X$ be the random variable that expresses the number of phone calls/hour, which means that $X$ is a discrete random variable, thus we are going to use the Poisson distribution and our $EX=\lambda = 4$ (calls/hour). The number of calls we expect in half an hour is $\mu = \dfrac{\lambda}2 = \dfrac{4}{2} = 2$. We know that the probability to receive $k$ calls during a $1/2$ (half) hour is:

$$\Pr(X = k) = e^{-\mu}\cdot \frac{\mu^k}{k!}.$$ Thus, the probability the operator misses that call in that time is: $$\Pr(X=1) = e^{-2}\cdot \frac{2^1}{1!}=\frac2{e^2}.$$

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    $\begingroup$ I think if there are more than one call in the half hour, then the operator still "misses a phone call". Thus you actually want $P(X \geq 1)$. But of course this can be calculated in more or less the same manner. $\endgroup$
    – Ian
    Sep 23, 2015 at 0:51
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    $\begingroup$ @Ian Yes, sure! But I think if it was the case you said, then it would be explicitly mentioned i.e. what is the probability that the operator misses at least one call? $\endgroup$
    – thanasissdr
    Sep 23, 2015 at 0:53
  • $\begingroup$ It would be clearer with the phrase "at least", but I think the spirit of the scenario is as I described. It is a little bit ambiguous, I agree. $\endgroup$
    – Ian
    Sep 23, 2015 at 0:57
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    $\begingroup$ It depends on the context. If you need to calculate the probability the operator misses exactly one call, then you need to calculate $P(X=1)$. If you need to calculate the probability the operator misses at least one call, then you need to calculate $P(X \ge 1)$. $P(X\le 1)$ is the probability the operator misses at most one call. $\endgroup$
    – thanasissdr
    Sep 23, 2015 at 1:16
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    $\begingroup$ "Did you miss a call?" "I didn't miss a call, boss." $\endgroup$
    – Christopher King
    Sep 23, 2015 at 1:51

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