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Homework Help: Probability - random variables, poisson/binomial distributions

  1. Jan 17, 2010 #1
    1. The problem statement, all variables and given/known data
    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.


    2. Relevant equations
    p_x(k) = (a^k)(e^-a)/(k!), poisson
    p_x(k) = (n choose k)(p^k)(q^{n-k}), binomial


    3. The attempt at a solution
    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:
    E(x) = np = 1000 x 0.001 = 1

    I don't really think I'm tackling either of these problems in the correct way but don't know what else to do. Thanks for any help.
     
  2. jcsd
  3. Jan 17, 2010 #2
    I don't think you have to worry about expected values here. When you model the number of errors with a binomial distribution, you want to find the probability of 0 "successes." "Success" in this case means a character being corrupted, which happens with probability .001. So you want to find p(k=0) using the formula

    [tex] p(k)={n \choose k} p^k (1-p)^{n-k} [/tex].

    For the poisson distribution, notice here that n is large and p small. When this is the case, a binomial distribution can be approximated by poission with parameter [tex]\lambda =np.[/tex]
     
  4. Jan 18, 2010 #3
    I'm still a bit confused, when I use poisson I get P(X=n) = (0.001^n)(e^-0.001)/(n!), so P(X=0) is e^-0.001 which is 0.99999 recurring. However, when I try it as you suggested, again using poisson but with np instead I get e^-1 which is about 0.37.
     
  5. Jan 23, 2010 #4
    Apologies for the (very) late response.

    When you write P(X=n) = (0.001^n)(e^-0.001)/(n!), you are saying that the Poisson parameter is .001, which is not true. In this problem, the Poisson parameter is supposed to be approximated by np, as mentioned above.

    The value .001 is used for modeling the number of errors as a binomial random variable, that is, the probability of no "successes" (i.e. no characters corrupted) is

    [tex] P(X=0)=p(0)={n\choose 0} .001^0 (1-.001)^{n-0} [/tex]

    where n=1000, the number of trials.
     
  6. Jan 23, 2010 #5
    Thanks, I understand now :)
     
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