# Queues with arrivals of random sized groups

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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In part (a), does arrivals refer to arrivals of groups or customers? That is, one arrival means one group arrives or one customer arrives? –  Dilip Sarwate Aug 30 '12 at 18:02
@DilipSarwate, that is a great observation, i assume customers, to be honest that is all the info provided in the question. –  Hardy Aug 30 '12 at 18:04

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. –  Dilip Sarwate Aug 30 '12 at 18:41
Thanks for spotting that. –  Robert Israel Aug 30 '12 at 20:05

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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i was wondering where does the the number 5 come. I assume to evaluate the answers a differentiation has been carried out,If you could please shed some light one why and when do we need to do that i would be very great full. –  Hardy Aug 30 '12 at 18:55
I assume to evaluate the answers a differentiation has been carried out... ?? // where does the the number 5 come... ??? Hrmm... you do know the definition of a Poisson process, don't you? –  Did Aug 30 '12 at 18:57
i have read it but obviously failed to pick up on the detail related to the differentiation needed to compute this i'll try re-examine this, but if you gave me this detail briefly it would be much appreciated. –  Hardy Aug 30 '12 at 19:05
it would be much appreciated... Obviously not. –  Did Aug 30 '12 at 23:22
@Hardy Didier gave the full calculation that you would need to do to follow up on Robert Israel's hint for the second question. Your allegation that Didier's answer is not a close fit to what was asked, and your previous comments about where does the number $5$ come and differentiation has been carried out indicates that you have a lot of learning to do. very great full indeed! –  Dilip Sarwate Aug 31 '12 at 11:36

\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*}