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Poisson vs Binomial approaches yield different results

  1. Jan 22, 2014 #1
    1. The problem statement, all variables and given/known data

    I made this question for myself to try to see if I could use two approaches (Poisson Distribution and Binomial Distribution) to solve a problem:

    Someone's average is to make 1 out of every 3 basketball shots.
    What are the chances she makes exactly 2 shots in a trial of 3 attempts?

    2. Relevant equations

    Poisson Distribution:
    (λ^k)(e^-λ)/(k!)

    Binomial Distribution:
    (n choose k)(P^k)((1-P)^(n-k))

    3. The attempt at a solution

    Poisson approach:
    λ is the average of successes in a given series length = 1
    k is the queried amount of successes in the same series length = 2
    Answer = (1^2)(e^-1)/(2!) = .184

    Binomial approach:
    n is the number of attempts = 3
    k is the number of queried successes = 2
    P is the probability of success = (1/3)
    Answer = (3 choose 2) (1/3)^2 (2/3)^1 = .222

    I have a feeling using Poisson here is wrong but I am not sure why. I have seen this type of problem used as an example for use of both Poisson and Binomial distributions.

    Thanks.
     
  2. jcsd
  3. Jan 22, 2014 #2

    Ray Vickson

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    The Poisson is wrong because it is not at all designed for this type of problem. Basically, the Poisson distribution is for counting events that can occur at any time, but when we want to know the probabilities of the number occurring in a particular interval. Your problem is different: events are "discrete"---think of them as occurring at times 1,2 and 3, for example. Under certain circumstances the Poisson can be used as an approximation to the Binomial which may be much easier to work with, but the circumstances where that holds are not present in your problem. I suggest you do a Google search on "Poisson distribution" for more information.
     
  4. Jan 22, 2014 #3

    haruspex

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    In a Poisson process, there's no limit to the number of successes you could get. In the OP problem, there cannot be more than three successes. So modelling it as Poisson means you will likely underestimate the probabilities of 0, 1, 2 or 3 successes, to balance the overestimation of more than 3.
     
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