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Binomial Distribution

  1. Nov 12, 2008 #1
    1. The problem statement, all variables and given/known data
    I am trying to figure out if I have a 20% chance to get what I want (let it = x) and I have 6 chances to do so (n=6), I am curious how I set this question up to find out my chances of getting 'x' once out of the 6 times I try.

    2. Relevant equations
    Binomial Distribution.

    x! / y!(x - y)!


    3. The attempt at a solution

    x = 0.2
    y = 0.8

    = 0.8! / 0.2! (0.8 - 0.2)!
    = 40320 / 1440
    = 28

    Any help would be appreciated.
     
  2. jcsd
  3. Nov 12, 2008 #2

    Dick

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    Science Advisor
    Homework Helper

    I don't think you quite get the binomial distribution. What is '.8!' suppose to mean? You might want to review it. On the other hand, the easy way to solve this problem is to figure out your odds of losing 6 straight times, call it P. Then your odds of winning are 1-P.
     
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