Probability Mass Function of Binomial Distribution

Theorem
The probability mass function (pmf) of a Binomial distributed random variable $$X$$ is equal to:


 * $$\Pr \left({X=x}\right) = \binom n x p^x(1-p)^{n-x} $$

where $$n$$ is the number of trials and $$p$$ is the probability of success.

Proof
Let $$B_i: \ i=1, 2, \ldots, \binom n x$$ be events such that:


 * (i) $$B_i$$ is the $$i^{th}$$ possible way to see $$x$$ successes in $$n$$ Bernoulli trials


 * (ii) $$\forall \ i \ne j: B_i \cap B_j = \varnothing$$

We can see that:


 * $$ \forall i: \Pr \left({B_i}\right) = p^x \left({1-p}\right)^{n-x} $$

This is true since there will be $$x$$ successes, each with probability $$p$$ of occurring, and $$n-x$$ failures each with probability $$1-p$$ of occurring.

Furthermore we can assume independent trials and thus the result follows.

See Bernoulli Process as Binomial Distribution for further analysis of this.

Now our task becomes finding:


 * $$\Pr \left({X=x}\right) = \Pr \left({\bigcup_{i=1}^{\binom n x} B_i}\right)$$ which is the probability of one of the $$\binom n x$$ outcomes occurring.

Then by the Inclusion-Exclusion Principle considered as an extension of the Addition Law of Probability we have that for any countable union of events:


 * $$\Pr \left({\bigcup_{i=1}^{n} A_i}\right) = \sum_{i=1}^{n} \Pr \left({A_i}\right) - \sum_{i \ne j: i, j=1}^{n} \Pr \left({A_i \cap A_j}\right) - \Pr \left({\bigcap_{i=1}^{n}A_i}\right)$$

Fortunately in this case the above reduces to:


 * $$\Pr \left({\bigcup_{i=1}^{n} A_i}\right) = \sum_{i=1}^{n} \Pr \left({A_i}\right)$$

Since the events are pairwise disjoint and $$\Pr(\varnothing) = 0$$

Thus we have:

$$ $$ $$ $$