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Cumulative distribution of binomial random variables

  1. Aug 5, 2012 #1
    1. The problem statement, all variables and given/known data
    The probability of being dealt a full house is approximately 0.0014. Find the probability that in 1000 hands of poker you will be dealt at least 2 full houses


    2. Relevant equations
    I can use binomial distribution.

    3. The attempt at a solution
    The probability of getting exactly two full houses would be (1000 choose 2) * (0.0014)^2 * (1 - 0.0014)^ 998

    However, because this problem asks for probability of being dealt *at least* two hands, I have to add the probabilities (1000 choose k) * (0.0014)^k * (1 - 0.0014)^ (1000-k), where k is any integers [2, 1000]

    Most problems in the textbook (Ross Probability) deal with situations where the possible values of k are only 4-6 integers and they can calculate it by hand, but I think I need to use summation for this problem. How do I do this kind of problems?
     
  2. jcsd
  3. Aug 5, 2012 #2
    Consider the Poisson distribution.
     
  4. Aug 5, 2012 #3

    Ray Vickson

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    Sometimes when it is too difficult to compute a probability P{A} directly, we instead compute the probability P{B} of the complementary event, then use P{A} = 1 - P{B}.

    RGV
     
  5. Aug 6, 2012 #4
    Ah I haven't thought of it that way. Thanks!
     
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