BINOMIAL DISTRIBUTION
DEFINITION
The Binomial distribution is the distribution of a sequence of independent Bernoulli variables that are identically distributed. While a Bernoulli variable measures the probability of success in one trial, the Binomial variable measures the probability of one or more successes in ‘$$n$$’ trials.
The Binomial distribution is therefore the sum of $$n$$ Independent and Identically Distributed (IID) Bernoulli trials. The distribution has the parameters $$n$$ and $$\theta$$, which indicate the number of trials and probability of success respectively. $$p$$ and $$q$$ are often used in place of $$\theta$$ and $$1-\theta$$ respectively

CALCULATOR

#### Enter the following details:

Total number of trials:

Number of successes required:

Probability of success for each trial:

Probability of successes:

Probability of or less successes:

Probability of more than successes:

Mean of the specified binomial distribution:

Variance of the specified binomial distribution:

Probability generating function:

Moment generating function:

Cumulant generating function:

Probability distribution
 x p(X=x) P(X ≤ x)
FORMULA AND DERIVATION
Probability distribution:-

Let us consider an example. A missile is fired, and the probability that it hits the target is 0.6. This can be modelled using a Bernoulli distribution with the probability of success $$p=0.6$$ and probability of failure $$q=0.4$$ .
Now consider a case where four missiles are fired one after the other. For each missile, the probability of hitting the target is the same$$(p=0.6)$$ , i.e. each missile has a Bernoulli distribution identical to the other missiles. And also, the probability of a missile hitting or missing the target is independent of the other missiles hitting or missing the target. We now need to find out the probability that two missiles out of four will hit the target. This can happen in various different orders. The first two missiles might hit and the last two might miss, or the first two might miss and the last two might hit, or it could be any other order. If S represents success(hit) and F represents failure(miss), the following will be the various orders in which two missiles could hit the target:-

SSFF
SFSF
SFFS
FSFS
FSSF
FFSS

So, there are six possible ways in which two out of four missiles can hit the target. This can be easily calculated using combinations, as $${{}^4C}_2$$ .
${{}^4C}_2=\frac{4!}{2!\left(4-2\right)!}=6$
Now, let us measure the probability of each of the six orders.

Orders Probability
SSFF $$0.6\times 0.6\times 0.4\times 0.4={0.6}^2{0.4}^2$$
SFSF $$0.6\times 0.4\times 0.6\times 0.4={0.6}^2{0.4}^2$$
SFFS $$0.6\times 0.4\times 0.4\times 0.6={0.6}^2{0.4}^2$$
FSSF $$0.4\times 0.6\times 0.6\times 0.4={0.6}^2{0.4}^2$$
FSFS $$0.4\times 0.6\times 0.4\times 0.6={0.6}^2{0.4}^2$$
FFSS $$0.4\times 0.4\times 0.6\times 0.6={0.6}^2{0.4}^2$$

So, the total probability of two hits and two failures is $$6\times {0.6}^2\times {0.4}^2$$. This can also be written as follows:-
${{}^4C}_2\times {0.6}^2\times {0.4}^2$
Therefore, the probability function of the binomial distribution can be written as follows:-
$P\left(X=x\right)={{}^nC}_x\times p^x\times q^{n-x}$
Mean:-

Since the binomial distribution is the sum of $$n$$ Bernoulli distributions, the binomial mean is $$n$$ times the Bernoulli mean.
$E\left(X\right)=\mu =np$
Variance:-

The binomial variance is $$n$$ times the Bernoulli variance.
$Var\left(X\right)={\sigma }^2=npq$
Probability Generating Function(PGF):-

The PGF for a Binomial distribution is
$G_X\left(t\right)={\left(pt+q\right)}^n$
Moment Generating Function(MGF):-

The MGF for a Binomial distribution is
$M_X\left(t\right)={\left({pe}^t+q\right)}^n$
Cumulant Generating Function(CGF):-

The CGF for a binomial distribution is
$C_X\left(t\right)=n\times {\mathrm{ln} \left({pe}^t+q\right)\ }$
EXAMPLES
Example 1

In a T-20 cricket match, the probability that a batsman of a particular team hits a ball for six, in the last over of the match, is 0.35. The supporters of the batting team wish to know the probability that the batsman will hit three sixes in the last over.

Here, each ball can be considered as a Bernoulli trial with probability of success equal to 0.35(hitting a six is considered as a success). So, 6 balls in an over form a series of 6 IID Bernoulli trials, which means that, number of successes in 6 trials can be modelled by the binomial distribution.

Here,
$$p=0.35$$
$$q=1-0.35=0.65$$
$$n=6$$
$$x=3$$(number of successes required)
$P\left(X=x\right)={{}^nC}_x\times p^x\times q^{n-x}$ $P\left(X=3\right)={{}^6C}_3\times {0.35}^3\times {0.65}^{6-3}$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{}^6C}_3\times {0.35}^3\times {0.65}^3$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =0.2355$
Example 2

A student wishes to attend an entrance exam, to get an admission in his favourite college. The exam consists of five questions. Each question is equally difficult and the probability that this student can get the right answer for a question is 0.75. The student would pass the exam if he answers at least three questions correctly. The student wishes to know the probability that he will answer three or more questions correctly and pass the exam.

In this case, each question can be considered as a Bernoulli trial with probability of success equal to 0.75 (getting the right answer is considered success). So, the 5 questions in the exam form a series of 5 IID Bernoulli trials, which means that, number of successes in 5 trials can be modelled by the binomial distribution.

Here,
$$p=0.75$$
$$q=1-0.75=0.25$$
$$n=5$$

Since the student needs a minimum of three right answers to pass the exam, the probability that he passes is the probability that he gets 3 right answers or 4 right answers or 5 right answers. Therefore, the required probability is
$P\left(X=3\right)+P\left(X=4\right)+P\left(X=5\right)$

$P\left(X=3\right)={{}^5C}_3\times {0.75}^3\times {0.25}^2$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =0.26387$
$P\left(X=4\right)={{}^5C}_4\times {0.75}^4\times {0.25}^1$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =0.39551$
$P\left(X=5\right)={{}^5C}_5\times {0.75}^5\times {0.25}^0$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =0.2373$
Therefore, the probability that the student will pass the exam is
$$0.26387+0.39551+0.2373=0.89648$$

Mean:-
$E\left(X\right)=\mu =np$ $\mu =5\times 0.75=3.75$
Variance:-
$Var\left(X\right)={\sigma }^2=npq$ ${\sigma }^2=5\times 0.75\times 0.25$ $\ \ \ \ =0.9375$