Understanding Binomial Distribution: Sum Always Equals 1?

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SUMMARY

The discussion clarifies that the sum of binomial probabilities from i=0 to n equals 1 due to the binomial theorem. Specifically, the formula Ʃ(p)i(1-p)n-i·K(n,i) represents the expansion of (p + (1-p))^n, which simplifies to 1^n, confirming that the total probability of all outcomes in a binomial distribution is always 1. This fundamental property is derived from the coefficients of the binomial expansion, ensuring the completeness of probability across all possible outcomes.

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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\cdotK(n,i)
Why is this sum always equal to 1?
 
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Hi, 4a,
straight out of the binomial theorem,\sum_{i=0}^n {\binom n i} p^i (1-p)^{n-i} = (p + (1-p))^n = 1^n = 1
 
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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