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Probability and Poisson Random Variable

  1. Feb 25, 2013 #1
    1. The problem statement, all variables and given/known data

    A trial consists of throwing two dice. The result is counted as successful if the sum of
    the outcomes is 12. What is the probability that the number of successes in 36 such trials
    is greater than one? What is this probability if we approximate its value using the Poisson
    random variable?

    3. The attempt at a solution

    So there is a 1/36 chance that throwing two dice results in a sum of 12. I'm not sure where to go from here though.
     
  2. jcsd
  3. Feb 25, 2013 #2
    The trials are independent since you are doing each trial separately. What do you know about the probability of independent events?
     
  4. Feb 25, 2013 #3
    P(EF) = P(E)P(F), so i guess i'd say E represents the event that the sum of the dice is 12 and F is the event that more than one trial is a success?
     
  5. Feb 25, 2013 #4

    Ray Vickson

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    Homework Helper

    Until you understand the basics, you should work with a simpler case.

    So, to start: suppose you do two independent trials. What is the probability that neither trial gives a '12'? What is the probability that both trials give a '12'? What is the probability that exactly one of the two trials gives a '12'?

    Now think about what would happen if you had three independent trials, then four independent trials, etc.
     
  6. Feb 25, 2013 #5
    Bleh so apparently i need to be using a binomial distribution on this problem.

    Inline8.gif

    So i have

    N = number of trials
    p = probability of a success in a given trial
    n = number of expected successful trials
    q = 1 - p

    N = 36,
    p = 1/36,
    1 = 35/36,
    n >= 1

    The problem is that i need an actual value for n, and n could be any number larger than 1 so it's throwing me off. How would i calculate this?
     
  7. Feb 26, 2013 #6

    Ray Vickson

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    The question asked for n > 1, not n >= 1. It makes a big difference. Anyway, you want the probability of the event {n ≥ 2} = {n = 2 or n = 3 or n = 4 or .... or n = 36}.
     
  8. Feb 28, 2013 #7
    By my calculations, n = 1 - [q^N] - [N*p*q^(N-1)]

    which evaluates to be 0.2642.
     
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