The Bernoulli and Binomial Distributions

The simplest experiments involving independence are repeated trials, each with just two outcomes: success (probability $p$) and failure (probability $1-p$). We encode success as $1$ and failure as $0$. Repeating the trial $n$ times, an outcome is a binary string $\omega=(s_1,\dots,s_n)$ with each $s_i\in\{0,1\}$, so the sample space is $\Omega=\{(s_1,\dots,s_n):s_i\in\{0,1\}\}$, of size $2^n$. Crucially we assume the trials are independent, which makes the probability of a whole string multiply across the individual trials: an outcome with some $1$'s and the rest $0$'s has probability $\mathbb{P}\{\omega\}=p^{\,(\#\,\text{of }1\text{'s})}\,(1-p)^{\,(\#\,\text{of }0\text{'s})}.$ For instance, the specific string $(0,1,0,0,1,0)$ over six trials — two successes and four failures — has probability $p^2(1-p)^4$. Notice this depends only on how many successes occurred, not on where they fell: every string with the same number of $1$'s is equally likely. That single fact is the engine behind everything that follows.

A single trial with two outcomes is modelled by the Bernoulli distribution. For a parameter $0\le p\le 1$, a random variable $X$ is Bernoulli with success probability $p$ if it takes only the values $X\in\{0,1\}$ with $\mathbb{P}(X=1)=p,\qquad \mathbb{P}(X=0)=1-p,$ and we write $X\sim\mathrm{Ber}(p)$. Here $X$ is the indicator of success: $X=1$ when the trial succeeds and $X=0$ when it fails. A fair coin's heads/tails is $\mathrm{Ber}(\tfrac12)$; rolling a six on a fair die is $\mathrm{Ber}(\tfrac16)$. The Bernoulli distribution is the atom of the theory: a sequence of $n$ independent trials gives rise to $n$ independent Bernoulli random variables $X_1,\dots,X_n$, all sharing the same parameter $p$. By independence, a joint outcome such as $\mathbb{P}(X_1=0,X_2=1,X_3=0,X_4=0,X_5=1,X_6=0)=p^2(1-p)^4$ — exactly the product structure of the previous concept.

Given $n$ independent Bernoulli trials $X_1,\dots,X_n$, the natural question is how many succeeded. Define $S_n=X_1+X_2+\cdots+X_n,$ which counts the number of $1$'s (successes) in the sample. To find $\mathbb{P}(S_n=k)$, note that each particular string with exactly $k$ successes has probability $p^k(1-p)^{n-k}$ (the product rule from independence). How many such strings are there? Choosing which $k$ of the $n$ positions are the successes is a combination, so there are $\binom{n}{k}$ of them. Since these strings are disjoint events of equal probability, we add them: $\mathbb{P}(S_n=k)=\binom{n}{k}p^k(1-p)^{n-k}.$ The two ingredients are worth separating: $p^k(1-p)^{n-k}$ is the probability of one way to get $k$ successes, while $\binom{n}{k}$ counts how many ways there are. Their product is the binomial probability.

The distribution of $S_n$ is the Binomial. For a positive integer $n$ and $0\le p\le 1$, a random variable $X$ has the binomial distribution with parameters $n$ and $p$ if $\mathbb{P}(X=k)=\binom{n}{k}p^k(1-p)^{n-k},\qquad k=0,1,\dots,n,$ and we write $X\sim\mathrm{Bin}(n,p)$. The two parameters carry plain meanings: $n$ is the number of trials and $p$ is the per-trial success probability. These values form a genuine distribution because they sum to $1$ — directly by the binomial theorem, $\sum_{k=0}^{n}\binom{n}{k}p^k(1-p)^{n-k}=\bigl(p+(1-p)\bigr)^n=1^n=1.$ A Bernoulli variable is just the special case $n=1$: $\mathrm{Bin}(1,p)=\mathrm{Ber}(p)$. To compute a cumulative probability $\mathbb{P}(X\le k)$ you sum the individual terms from $0$ to $k$; for the tail $\mathbb{P}(X\ge 1)$ the complement $1-\mathbb{P}(X=0)=1-(1-p)^n$ is almost always quicker.