# 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?

• Are you possibly asking why the binomial theorem works? The answer given below demonstrates how it works, if you understand the binomial theorem. – WaveX Sep 25 '18 at 0:53
• You can prove the binomial theorem by induction. Assume the expression is valid for $n$ and show that it is also valid for $n+1$, in which $n$ is replaced by $n+1$ in the summation. Finally, you show that it is valid for $n=1$. Here's a nice detailed proof: math.stackexchange.com/questions/2205908/… – minmax Sep 25 '18 at 1:20

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.

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.