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Probability: Poisson distribution involving customer arrivals

  1. Feb 17, 2014 #1
    1. The problem statement, all variables and given/known data

    There are two stores A and B.

    Customers can equally enter one of the two stores, i.e., for a specific customer, the probabilities she enters store A or B both are 0.5.

    If the total number of customers in two stores has the Poisson distribution of parameter λ, then

    Question 1: What is the probability that the number of customers in store A is non-zero and store B has no customers;

    Question 2: What is the probability that the number of customers in store A is exactly 2 and store B has no customers.


    2. Relevant equations

    Poisson distribution: p(x)=λx/x! * e

    3. The attempt at a solution

    Answer 1: (1-p(0))*(0.5+0.52+0.53+...)=1-e
    Comment: The probability that there are some customer in some store is 1 - p(0), then the probability that x customers entered store A is 0.5x, hence their product should yield the desired answer?

    Answer 2: p(2)*0.52
    Comment: Similar strategy as answer 1.

    Are these correct?

    Thank you in advance!
     
    Last edited: Feb 17, 2014
  2. jcsd
  3. Feb 17, 2014 #2

    Ray Vickson

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    You answer to (2) is correct, but your answer to (1) is not.
     
  4. Feb 17, 2014 #3
    Ray Vickson, thank you for your replay.

    If answer (2) is correct, then answer (1) is p(1)*0.5 + p(2)*0.52 + p(3)*0.53 + ...

    Is this correct?

    If it is, then is there a concise form of the summation?
     
  5. Feb 17, 2014 #4

    Ray Vickson

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    Yes, and yes. For the latter, see https://www.efunda.com/math/exp_log/series_exp.cfm .
     
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