# Binomial distribution

1. Feb 19, 2012

### aaaa202

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$\cdot$K(n,i)
Why is this sum always equal to 1?

2. Feb 19, 2012

### dodo

Hi, 4a,
straight out of the binomial theorem,$$\sum_{i=0}^n {\binom n i} p^i (1-p)^{n-i} = (p + (1-p))^n = 1^n = 1$$

3. Feb 19, 2012

### alan2

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.