Binomial Distribution

As we know, binomial distribution has its importance at the top of the hierarchy of subjects. Its application is widely extended all over the statistical procedures. 

It represents the distribution where there is two possibilities (one is success and one is failure) which can be describes as the Bernoulli distribution.

\begin{equation} P=Success\;\;\; 1-P=Failure\end{equation}

And, additional to this information, binomial distribution has also a feature that allows us to do experiment of this Bernoulli distribution for repeated several times and repetitions, which are considered to be independent action. 

Think of a coin (which is the most simple way to understand the binomial distribution). You want to toss a coin for 3 times and expect to get exactly 2 heads. 

We know that for these three coins, having head or tail is independent of each other. And also know that probability of getting head for tossing a coin is 0.5. 

Firstly, we will examine it in a clustering method. Because we want to have 2 heads for 3 coins we can have:

\begin{equation}[H,H,T]\;\;\;[H,T,H]\;\;\;[T,H,H]\end{equation}

Then, we can calculate the probabilities for each of these three cases.

\begin{equation}[H,H,T]=\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}=\frac{1}{8}\end{equation}

And because we have 2 remaining cases, we have to also add these probabilities of these cases.

\begin{equation}3\times\frac{1}{8}=\frac{3}{8}\end{equation}

As you can observe, we find the probability of a one case and multiply the number of combinations of the required distribution.

So we can explain the equation for binomial distribution as:

\begin{equation}C^n_{x}\times P^x\times(1-P)^{(n-x)}\end{equation}

In this case x represents the number of success (As you can infer, our coin example the number of success 2, in which we require 2 heads). And n is the number of trial (Which is 3 for our coin example, where we tossed our coin 3 times). And P is the probability of success (Where in every trial the probability of having head is 0.5). And we multiply with the number of combinations we have.

For a final point, let's solve a problem.

A test of performence of new hired employees has been studied. And it has been found that 0.85 of employees quit the job after a week later. Find the probability that at least 2 of the 7 employees quit the job after a week later.

\begin{equation}n=7\;\;\;P=0.85\;\;\;x\ge2\end{equation}

\begin{equation}p(x\ge2)=1-p(x=0)-p(x=1)\end{equation}

\begin{equation}p(x=0)=C^7_{0}\times(0.85)^0\times(0.15)^7\end{equation}

\begin{equation}p(x=1)=C^7_{1}\times(0.85)^1\times(0.15)^6\end{equation}

With the help of R we can find:

pbinom(1,size = 7,prob = 0.85,lower.tail = FALSE)