Binomial distribution

Binomial distribution
$X\sim\text{Bin}(n,p)$
$p\in$$[0,1]$$\subseteq\mathbb{R}$, $n\in\mathbb{N}_{\ge 0}$
Definition
p.m.f	$\mathbb{P}[X\eq k]:\eq{}^n\text{C}_k\ p^k(1-p)^{n-k}$

Derivation

Let there be $n$ items, $X_1,\ldots,X_n$ which can take the values "yes" or "no", we denote by $y_i$ the $i^\text{th}$ $X_j$ taking the value $y$ and similarly for $n_j$.

We are interested in getting exactly $k$ $y$s. (specifically how many combinations there are of various $X_i$ for getting $y$s, we consider for example $y_1y_2$ the same as $y_2y_1$, these are distinct permutations but the same combinations)

• We can choose the first $y$ to come from any of the $n$ $X_j$s so there are $n$ ways to choose this
• The next $y$ can only come from $n-1$ $X$s (as 1 of them is already taken)
• The next $y$ can only come from $n-2$ $X$s

and so on, the result is $n(n-1)(n-2)\cdots(n-(k-1))$, we may write this as $\frac{n!}{(n-k)!}$

Next consider the case where $k\eq 3$, then one of the ways to choose $k$ may be $(1,2,4)$ (meaning $y_1y_2y_4$ the rest $n$s), and another way is $(2,1,4)$, meaning $y_2y_1y_4$), notice that these are "the same" (as combinations, but are different permutations), they both have $X_1$, $X_2$ and $X_4$ being "yes".

Notice that choosing from $\{1,2,4\}$ there are 3 ways to choose the first element, 2 ways for the second and only one way (the remaining item) for the third.

Thus, in general, of the $k$ chosen $X$s to represent "yes"s, there are:

• $k$ ways to choose the first element
• $k-1$ ways to choose the second element
• ....
• $2$ ways to choose the $k-1$th element,
• $1$ way to choose the final element.

So there are $k(k-1)\cdots 2\cdot 1$ ways to choose these $k$ chosen elements.

This is just $k!$

Thus the $\frac{n!}{(n-k)!}$ ways to choose count the same collection of $k$ elements $k!$ times (as in the example, the elements 1, 2 and 4 were counted 6 times, 2 were explicitly given)

So we divide $\frac{n!}{(n-k)!}$ by $k!$ to yield

• $\frac{n!}{k!(n-k)!}$ this is just ${}^n\text{C}_k$

Motivating example

Suppose we flip a coin 3 times and count the number of heads, we want to find the probability that there is exactly 1 head. Notice that there are 3 distinct outcomes which have exactly one head, htt, tht and tth, assuming each instance of one head and two tails has the same probability ($p(1-p)(1-p)$ for independent flips, and $p$ the probability of a head on a single flip) we can just multiply that probability by 3.

In general for $n$ coin flips, the probability of exactly $k$ heads depends ${}^n\text{C}_k$ distinct outcomes, each with $k$ heads.

As above, assuming each flip is independent and the probability of any individual flip yielding heads being $p$, then the probability of exactly $k$ heads from the $n$ flips is:

• ${}^n\text{C}_k\ p^k(1-p)^{n-k}$