Binomial Distribution equals 1 Proof In order to prove $b(x,n,p) =$ ${n \choose x}p^x(1-p)^{n-x}$$for$ $x=0,1,2,...,n$ (and $0$ otherwise) we must show $1 = \sum_{x=0}^n{n \choose x}p^x(1-p)^{n-x}$
However, all the answers I've seen are just stating the binomial theorem with no work shown. Would anyone be willing to show me the steps for $1 = 1^n = (p+1-p)^n =  \sum_{x=0}^n{n \choose x}p^x(1-p)^{n-x}$ so I can grasp the idea better?
 A: One is given that 
$$ (x+y)^n = \sum_{i=0}^n {n \choose i} x^i y^{n-i} $$
This identity is called the Binomial. Now, if we put $y=1-x$ in our formula binomial we get
$$ 1=(x+1-x)^n = \sum_{i=0}^n {n \choose i} x^i (1-x)^{n-i} $$
If we put $x=p$ and $i=x$, then we get your result
Neymar.
A: Here is a probabilistic argument that you might find helpful:
Imagine you have a loaded coin which when flipped has probability $p$ of landing heads, and probability $q=1-p$ of landing tails. Suppose you flip this coin $n$ times. Let $E_k$ be the event that you get $k$ heads. Then the probability of $E_k$ is equal to the probability of getting a specific sequence of $k$ heads and $(n-k)$ tails (which is $p^kq^{n-k}$), times the number of such sequences (which is ${n \choose k}$, as we can have the $k$ heads occur in any of the $n$ flips). Thus the probability of $E_k$ is ${n \choose k} p^kq^{n-k}$.
Now you have to get some number of heads from $0$ to $n$, and the $E_k$ are all disjoint, so their probabilities must sum to $1$.
Note: This argument of course only works for $p\in[0,1]$. Technically, it then follows for all $p$ by analytic continuation, though it is much easier to just prove it with induction at that point. The above argument is more meant to provide some intuition for why the statement must be true.
