# Prove that Binomial Distribution is a discrete distribution?

I struggled to come up with an approach and I'm not sure if induction is needed. Please help!

EDIT:

Thanks for all the quick feedback!

Is there a way to prove that for a random variable x with a binomial distribution, P(x=0) + P(x=1) + ... + P(x=n) = 1? Possibly through induction?

• It is discrete by its very definition, there is nothing to be proven. – Yves Daoust Jun 6 at 7:26
• How can you use induction on non integer points? – Upstart Jun 6 at 7:29
• Write down the definition of a binomial distribution. Then write down your definition of a discrete distribution. The rest should be obvious. – Theoretical Economist Jun 6 at 7:49
• Thanks for all the quick feedback! Is there a way to prove that for a random variable x with a binomial distribution, P(x=0) + P(x=1) + ... + P(x=n) = 1? Possibly through induction? – Shukie Jun 6 at 8:01

A random variable X is said to have the Binomial distribution with parameters $$N,p$$ if

$$p(X =x) = \begin{cases} \binom{N}{x} \,\,p^x \,\,(1-p)^{N-x} & 0 \leq x \leq N \\\\ 0 & otherwise \end{cases}$$

since the support of random variable is $$\{0,1,2..,N\}$$.we can conclude that it is discrete distribution

EDIT:

following is the classic (and elegant) proof if you know Binomial expansion

$$= \sum_{x=0}^N \binom{N}{x} \, p^x \, (1-p)^{N-x}$$

$$=\Big( p + 1-p \Big)^N$$ ( By binomial expansion )

$$= 1^{N}$$

$$= 1$$

• oops sorry.edited :) – viru Jun 6 at 7:45
• Thanks for all the quick feedback! Is there a way to prove that for a random variable x with a binomial distribution, P(x=0) + P(x=1) + ... + P(x=n) = 1? Possibly through induction? – Shukie Jun 6 at 8:01
• I have added you can check now. – viru Jun 6 at 16:27

If $$X$$ has binomial distribution with parameters $$n$$ (and $$p$$) then $$P(X\in\{0,1,\dots,n\})=1$$.

The set $$\{0,1,\dots,n\}$$ is a countable set.

This together states that binomial distribution can be labeled as a discrete distribution.