Binomial distribution

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  • #1
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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?
 

Answers and Replies

  • #2
695
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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]
 
  • #3
323
56
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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