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Binomial distribution

  1. Feb 19, 2012 #1
    Is quite easy to understand. What I don't understand though is this:
    When you sum over all the binomial probabilities from i=0 to n you should get 1, as this corresponds to the total probability of getting any outcome. I just don't understand what it is, that guarantees that you always get one when you sum over:
    Ʃ(p)i(1-p)n-i[itex]\cdot[/itex]K(n,i)
    Why is this sum always equal to 1?
     
  2. jcsd
  3. Feb 19, 2012 #2
    Hi, 4a,
    straight out of the binomial theorem,[tex]\sum_{i=0}^n {\binom n i} p^i (1-p)^{n-i} = (p + (1-p))^n = 1^n = 1[/tex]
     
  4. Feb 19, 2012 #3
    The quick way is to note that the binomial coefficients are the coefficients of successive terms in the expansion of (x+y)^n (see binomial theorem). So your summation is the expansion of (p+(1-p))^n which is equal to 1 for any n.
     
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