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I am trying to solve the following problem.

Customers arrive at a service facility in groups. The groups arrive according to a homogeneous Poisson process at the rate of 5 per minute. The number N of individuals in each group is described by the distribution $P_N(1) =0.3$, $P_N(2) =0.6$ and $P_N(3) =0.1$

  1. What are the probabilities of zero, one and two arrivals in one minute?
  2. Suppose a customer in the system is chosen at random. What is the probability that he/she arrived in a group of size n?

Thoughts towards a solution $P(S) = 0.3s+0.6s^{2}+0.1s^{3}$ So the generating function would be $G(s) = e^{5t(0.3s+0.6s^{2}+0.1s^{3})}$

Firstly i am unsure how to use the generating function to evaluate the first part. I think the probability is given by the coefficients of the z terms so the answers to the three questions should be 0, $e^{0.15t}$ and $e^{3t}$ respectively.

As per the second part i suspect the answer would be obtained by a conditional probability, as in the product of he/she arriving in a group of size n(1/3), times the probability of such a group coming in to the store, as stated above (assuming that is correct) but i am unsure how to put this into a formula

Any help would be much appreciated.

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    @DilipSarwate, that is a great observation, i assume customers, to be honest that is all the info provided in the question.2012-08-30

3 Answers 3

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Answer to part (a):

$\begin{align*} P\{\text{no customers in one minute}\} &= P\{\text{no groups in one minute}\}\\ &= \exp(-5)\\ P\{\text{one customer in one minute}\} &= P\{\text{one group (with one customer) in one minute}\}\\ &= (5\exp(-5))\cdot 0.3\\ P\{\text{two customers in one minute}\} &= P\{\text{one group (with two customers) in one minute}\}\\ &\quad + P\{\text{two groups with one customer each in one minute}\}\\ &= (5\exp(-5))\cdot 0.6 + \frac{5^2}{2!}\exp(-5)\cdot (0.3)^2\\ \end{align*}$

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Zero customer means zero group, which happens with probability $\mathrm e^{-5}$. One customer means one group of size one, which happens with probability $\mathrm e^{-5}\cdot5\cdot0.3$. Two customers means either one group of size two, which happens with probability $\mathrm e^{-5}\cdot5\cdot0.6$, or two groups of size one, which happens with probability $\mathrm e^{-5}\cdot\frac{5^2}{2!}\cdot(0.3)^2$, for a total probability of $\mathrm e^{-5}\cdot5\cdot0.6+\mathrm e^{-5}\cdot\frac{5^2}{2!}\cdot(0.3)^2$.

In $m$ minutes with $m$ large, roughly $0.3m$ groups of size 1 and $0.6m$ groups of size 2 and $0.1m$ groups of size 3 arrived, for a total of $0.3m+2\cdot0.6m+3\cdot0.1m$ customers. Amongst these, roughly $2\cdot0.6m$ were from a group of size 1 hence the probability that a random customer comes from a group of size 1 is $2\cdot0.6/(0.3+2\cdot0.6+3\cdot0.1)=2/3$.

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    @Hardy As if I had said anything even remotely close to this... Enough (bis and last).2012-08-31
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Hint for (1): Zero arrivals is easy, one arrival means one group with one person; two arrivals means either one group with two persons or two groups of one each.

Hint for (2): Expected number of people in groups of size $n$, divided by expected number of people.

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    I corrected what appeared to be a minor typo. Please restore the original version if you don't agree with the change.2012-08-30