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Poisson Distribution problem

  1. Nov 8, 2008 #1
    1. The problem statement, all variables and given/known data
    The number of customers entering a cafe during tea time is known to be poisson distribution with λ = 5.
    on a particular day, given that at least 2 customers have entered the cafe during the tea time. what is the probability that at least 1 more customers will enter the cafe during tea time?

    3. The attempt at a solution
    I was thinking of using this :

    [tex]P(X \geq 1 | X \geq 2) = \frac{P(X \geq 1 \cap X \geq 2)}{P(X \geq 2)}[/tex]

    However, i soon found that if i do that, the Probability will go to 1. it does not make sense. Please advice. thank you.
  2. jcsd
  3. Nov 9, 2008 #2


    Staff: Mentor

    The conditional probability on the left side represents the probability that X >= 1, given that X >= 2. If X >= 2, then it will certainly be greater than 1. I think you want P(X >= 3 | X >= 2).
  4. Nov 9, 2008 #3
    Thanks for the reply. But I still don't quite get it. Any advise? Thanks..
  5. Nov 9, 2008 #4


    Staff: Mentor

    If you know that there are two people in the cafe, the probability that there is at least one person is 1. For a different example, if I see that you have a dollar, then I know for certain that you have 50 cents (you might have to change the dollar, though).
    You know that there are at least two people in the cafe. "one more customer" means that there are three people. "at least one more customer" means that there are three or more.
    What you're trying to find is the probability of 3 or more people in the cafe, given that there already two people there.
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