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Poisson and binomial distributions, corrupted characters in a file

  1. Jan 18, 2010 #1
    A text file contains 1000 characters. When the file is sent by email from one machine to another, each character (independent of other characters) has probability 0.001 of being corrupted. Use a poisson random variable to estimate the probability that the file is transferred with no errors. Compare this to the answer you get when modelling the number of errors as a binomial random variable.

    Poisson:
    Let x be the number of corrupted characters.
    E(X) = 0.001 = a
    P(X=n) = (0.001^n)(e^-0.001)/(n!)
    P(X=0) = e^-0.001

    Binomial:
    np = 1000 x 0.001 = 1
    When I approximate using poisson but with np instead I get e^-1 which is about 0.37.

    This doesn't seem right?
     
  2. jcsd
  3. Jan 18, 2010 #2

    Mark44

    Staff: Mentor

    Show us your calculation of P(X = 0) for the binomial case.
     
  4. Jan 18, 2010 #3
    I used poisson but approximated lambda as np, so ((np)^k)(e^-np)/k! where np = 1000x0.001 = 1 and k = 0, so we get (1^0)(e^-1)/0! = e^-1.

    I don't know how I could use the proper binomial random variable.
     
  5. Jan 18, 2010 #4

    Mark44

    Staff: Mentor

    I believe you are supposed to assume that X is a binomial r.v., where the probability for a given character being in error is .001. You want P(X=0). See this Wikipedia page for more information.
     
  6. Jan 18, 2010 #5
    Thanks, when I do it that way I get (1000 choose 0)(0.001^0)(1-0.001)^1000 = 0.3676...
    So, is it when I have calculated my poisson RV that I've gone wrong and I should have done this as I was previously trying to approximate binomial?
     
  7. Jan 18, 2010 #6
    Actually, have just read up some more on Poisson distributions and I know where I've gone wrong. Thanks for all your help!
     
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