(related to Proposition: Binomial Distribution)
Let \(X\) be the random variable counting the number \(k\) of the realizations of \(A\). Because we repeat the experiment \(n\) times, each time the event \(A\) will occur with the probability \(p\) and the complement event \(\overline{A}\) will occur with the probability \(1-p\). Imagine, in the first \(k\) repetitions, we observe only \(A\), while in the remaining \(n-k\) repetitions we observe only \(\overline{A}\). Then the joint event will be
\[\underbrace{A,\ldots,A}_{k\text{ times}}\underbrace{\overline{A},\ldots,\overline{A}}_{n-k\text{ times}}.\quad\quad ( * )\]
Because we conduct a Bernoulli experiment, all repetitions are mutually independent. Therefore, their joint probability of observing exactly \(( * )\) is the product of all probabilities of each realization, which gives us the formula
\[p^k(1-p)^{n-k},\quad\quad 0\le k\le n.\]
Because the binomial coefficient \(\binom nk\) is the number of possible words of length \(n\) with \(k\) letters \(A\) and \(n-k\) letters \(\overline{A}\), in which the order of letters does not play any role, the joint probability of observing exactly \(k\) times the event \(A\) is given by
\[p(X=k)=\binom nk p^k(1-p)^{n-k},\quad\quad 0\le k\le n.\]
For all other real numbers \(k\), we have the \(p(X=k)=0\). Therefore, the probability mass function is
\[p(X = k)=\begin{cases} \binom nk p^k(1-p)^{n-k}&\text{for }k=0,1,\ldots n\\\\ 0&\text{else.}\end{cases}\]
It follows that the probability distribution of the random variable \(X\) is given by
\[\begin{array}{rcll} p(X \le x)&=&0&\text{for }x < 0\\ p(X \le x)&=&\sum_{k=0}^{k=x}\binom nk p^k(1-p)^{n-k}&\text{for }0\le x < n\\ p(X \le x)&=&1&\text{for }x \ge n\\ \end{array}\]
