Discrete distributions

Suppose our random variable $X$ can take only values $v_1, v_2, \ldots, v_N$. We insist that $0\leq f_X(v_i) \leq 1$, and

$$\sum_{i=1}^N f_X(v_i) = 1.$$

If each $v_i$ is equally likely to occur, then these conditions lead to a uniform distribution with $f_X(v_i) == 1/N$. A binomial distribution expresses how often we expect a particular value to occur out of $n$ events. If $f_X(e)$ = $p$ (the probability that our chosen event occurs once is $p$), then the probability of exactly $k$ occurrences is:

$$f_X(k) = {n \choose k} p^k(1-p)^{n-k}.$$

The reasoning is that in a sequence of $n$ events there are $n$   choose$k$ ways to select exactly $k$ event $e$s, and the probability of each selection is the product of the $p$s with the product of the $(1-p)$s (the probability that $e$ doesn't occur).

This is a discrete normal distribution.