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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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    In part (a), does arrivals refer to arrivals of _groups_ or _customers_? That is, one arrival means one group arrives or one customer arrives?2012-08-30
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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

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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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    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.2012-08-30
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    *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?2012-08-30
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    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.2012-08-30
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    *it would be much appreciated*... Obviously not.2012-08-30
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    Although you did type more text and gave more details towards a solution for my problem, and i genuinely do appreciate that, but for the second part the other answer is much a closer fit to what was asked. Please do n't take this the wrong way. This decision is not based on the comment discussion above. Also on that note I am trying to self learn this stuff and the book subscribed from school seems to be rather dry and leaves many important thing under emphasized. Hence the request. Thanks for your help with the answer :-)2012-08-31
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    @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!2012-08-31
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    @Hardy All this is not very convincing, to put it mildly. As far as I am concerned, after you asked me for some more details and I inquired about your level of mathematical sophistication, I was rather surprised to see you accept immediately another, **less detailed** (although excellent), answer. This seems to make no sense at all but surely you have your reasons. By the way, surely this can only mean you are able to transform the hints in the accepted answer into a full solution, thus I feel free to consider your previous request as moot, hence to add no detail at all to my own post. Agreed?2012-08-31
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    @Hardy As an aside, I note that some parts of your previous comment are simply insulting my intelligence--so let us leave them aside.2012-08-31
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    @DilipSarwate I do indeed have a lot of learning to do, I am not ashamed to admit it in fact i am actually proud that i am learning. I am not interested in fighting. This place(forum) is about maths let's not let it become a high school yard.2012-08-31
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    @Did Usually i never ask for help on this site with homework problems. Actually the homework submission yesterday was the very first time. The second part of your answer calculates probability that a random customer comes from a group of size 1 and my question was for the nth case. I could not make the transition and submitted home work with the hint From Robert Israel's answer. So to me that appears to be a closer answer. As per my question in the comments, I waited for a while but you did not provide any details or clues and now you are trying to prove my request was a moot. (contd)2012-08-31
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    @Did Look all you had to say look up when a differentiation needs to be carried out on pgf if you could not be bothered with any details. I am not trying to insult you or your intelligence. If you think i did wrong by you i apologize.2012-08-31
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    @Hardy Thanks for having confirmed everything I said. All in all, the best part of your comments is when you shamelessly state that a full explanation of the principle behind question 2, plus the explicit computations of the case of size 1, were not sufficient to solve the cases of size 2 and size 3. Enough.2012-08-31
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    @did I can not be bothered with this discussion. So i would not be offering any statements any more. You proved that I am dumb. You win :-)2012-08-31
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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