Binomial Distribution With Examples


Probability / Tuesday, October 9th, 2018

Binomial Distribution

Before starting our discussion on Binomial Distribution we have to understand what Bernoulli Trials is.

Bernoulli Trials

Trials are called Bernoulli trials if they meet the following requirements:
1. Each trial has only two possible outcomes called success and failure.
2. The probability of the outcome of any trial remains fixed.
3. All the trials are statistically independent.

Binomial Distribution Definition

Let the random variable X be defined as follows:
\[\text{X }=\text{ Number of successes}\]
The we call X a binomially distributed random variable with parameter n and p if its probability distribution is given by
\[p\left( X=x \right)={}^{n}{{C}_{x}}.{{p}^{x}}{{\left( 1-p \right)}^{n-x}},x=0,1,….n\]
Where p = the probability of Success
And (1 – p) = q = the probability of Failure
\[\text{We write X}\sim B\left( n,p \right)\]

Conditions for a Binomial Random Variable

1. The trials must be Bernoulli trials in that the outcomes can only be either success or failure.
2. The outcomes of the trials must be independent.
3. The probability of success in each trial must be constant.

Mean, Variance and Mode in Binomial Distribution

Mean
\[=\sum\limits_{x=1}^{n}{x.}\frac{n!}{\left( n-x \right)!x!}{{p}^{x}}{{q}^{n-x}}\]
\[=np\sum\limits_{x=1}^{n}{\frac{\left( n-1 \right)!}{\left( n-x \right)!\left( x-1 \right)!}{{p}^{x-1}}{{q}^{n-x}}}\]
\[\therefore E\left( X \right)=np\]

Variance
\[V\left( X \right)=E\left( {{X}^{2}} \right)-{{\left[ E\left( X \right) \right]}^{2}}\]
\[E\left( {{X}^{2}} \right)=\sum\limits_{x=1}^{n}{{{x}^{2}}p\left( x \right)}=\sum\limits_{x=1}^{n}{{{x}^{2}}.{}^{n}{{C}_{x}}.{{p}^{x}}{{q}^{n-x}}}\]
\[=\sum\limits_{x=1}^{n}{{{x}^{2}}\frac{n!}{\left( n-x \right)!x!}{{p}^{x}}{{q}^{n-x}}}\]
\[=\sum\limits_{x=1}^{n}{\frac{x\left( x-1 \right)n!}{\left( n-x \right)!x!}{{p}^{x}}{{q}^{n-x}}}+\sum\limits_{x=1}^{n}{x\frac{n!}{\left( n-x \right)!x!}{{p}^{x}}{{q}^{n-x}}}\]
\[=n\left( n-1 \right){{p}^{2}}\sum\limits_{x=2}^{n}{\frac{\left( n-2 \right)!}{\left( n-x \right)!\left( x-2 \right)!}{{p}^{x-2}}{{q}^{n-x}}}+np\]
\[\Rightarrow E\left( X \right)=n\left( n-1 \right){{p}^{2}}+np\]
\[\Rightarrow E\left( X \right)={{n}^{2}}{{p}^{2}}-n{{p}^{2}}+np\]
\[\therefore V\left( X \right)={{n}^{2}}{{p}^{2}}-n{{p}^{2}}+np-{{n}^{2}}{{p}^{2}}\]
\[\therefore V\left( X \right)=np\left( 1-p \right)=npq\]
\[\therefore \text{Standard Deviation}=\sqrt{V\left( X \right)}=\sqrt{npq}\]

Mode

We have to find that value of the random variable X following Binomial distribution for which the probability of occurrence is maximum. Thus, if X = k is the modal value, its definition provides

Now two cases arise
Case1. Let (n+1)p be an integer then k can take two values namely k = (n+1)p and k =(n+1)p – 1. Thus there are two modes and the distribution is said to be binomial.
Case2. Let (n+1)p be a fraction then since k is an integer it will take only one value that is k = greatest integer less than (n+1)p and the distribution is unimodal.

Moments generating function (mgf)

Let X is a Binomial random variable with parameter n and p, its moment generating function is defined as:
\[{{M}_{X}}\left( t \right)=E\left( {{e}^{tX}} \right)=\sum\limits_{k=0}^{n}{{}^{n}{{C}_{k}}.{{p}^{k}}{{q}^{n-k}}.{{e}^{tk}}}\]
\[\Rightarrow {{M}_{X}}\left( t \right)=\sum\limits_{k=0}^{n}{{}^{n}{{C}_{k}}.{{\left( p{{e}^{t}} \right)}^{k}}{{q}^{n-k}}}={{\left( q+p{{e}^{t}} \right)}^{k}}\]

 Example 01

6 coins are tossed. Find the probability of getting (i) exactly 3 heads, (ii) at most 3 heads, (iii) at least 3 heads.

Solution:

Let X = Number of heads that appear.
Probability of getting a head, p = ½, n = 6
\[(i)P\left( \text{getting exactly 3 heads} \right)=p\left( X=3 \right)\]
\[={}^{6}{{C}_{3}}.{{\left( \frac{1}{2} \right)}^{3}}{{\left( 1-\frac{1}{2} \right)}^{6-3}}=\frac{5}{16}\]
\[(ii)P\left( \text{getting at most 3 heads} \right)=p\left( X\le 3 \right)\]
\[=p\left( X=0 \right)+p\left( X=1 \right)+p\left( X=2 \right)+p\left( X=3 \right)\]
\[={}^{6}{{C}_{0}}.{{\left( \frac{1}{2} \right)}^{0}}{{\left( 1-\frac{1}{2} \right)}^{6-0}}+{}^{6}{{C}_{1}}.{{\left( \frac{1}{2} \right)}^{1}}{{\left( 1-\frac{1}{2} \right)}^{6-1}}+{}^{6}{{C}_{2}}.{{\left( \frac{1}{2} \right)}^{2}}{{\left( 1-\frac{1}{2} \right)}^{6-2}}+{}^{6}{{C}_{3}}.{{\left( \frac{1}{2} \right)}^{3}}{{\left( 1-\frac{1}{2} \right)}^{6-3}}\]
\[=\frac{21}{32}\]
\[(iii)P\left( \text{getting at least 3 heads} \right)=p\left( X\ge 3 \right)\]
\[=1-p\left( X<3 \right)\]
\[=1-\left[ p\left( X=0 \right)+p\left( X=1 \right)+p\left( X=2 \right) \right]\]
\[=1-\left[ {}^{6}{{C}_{0}}.{{\left( \frac{1}{2} \right)}^{0}}{{\left( 1-\frac{1}{2} \right)}^{6-0}}+{}^{6}{{C}_{1}}.{{\left( \frac{1}{2} \right)}^{1}}{{\left( 1-\frac{1}{2} \right)}^{6-1}}+{}^{6}{{C}_{2}}.{{\left( \frac{1}{2} \right)}^{2}}{{\left( 1-\frac{1}{2} \right)}^{6-2}} \right]\]

 Example 02

Probability that a car driving the entire length of the certain turnpike will have a blowout is 0.05. Find the probability that among 17 cars travelling the length of turnpike (i) exactly one will have a blowout, (ii) 2 or more will have a blowout.

Solution:

Let X = Number of cars having a blowout.
Here, n = 17, p = 0.05, q = 1 – p =0.95
\[(i)P\left( \text{exactly one will have a blowout} \right)=p\left( X=1 \right)\]
\[={}^{17}{{C}_{1}}.{{p}^{1}}{{q}^{16}}={}^{17}{{C}_{1}}\left( 0.05 \right){{\left( 0.95 \right)}^{16}}\]
\[(ii)P\left( \text{2 or more will have a blowout} \right)=p\left( X\ge 2 \right)\]
\[=1-p\left( X<2 \right)=1-\left[ p\left( X=0 \right)+p\left( X=1 \right) \right]\]

 Example 03

A perfect die is tossed 100 times in sets of 8. The occurrence of a 5 or 6 is called a success. How many times do you expect to get 3 success?

Solution:
\[\text{P}\left( \text{Success} \right)\text{ }=\text{ P}\left( \text{getting 5 or 6} \right)=\frac{1}{6}+\frac{1}{6}=\frac{1}{3}\]
\[\therefore p=\frac{1}{3},q=\frac{2}{3},n=8\]
\[P\left( \text{getting 3 success} \right)=p\left( X=3 \right)\]
\[={}^{8}{{C}_{3}}.{{p}^{3}}{{q}^{5}}={}^{8}{{C}_{3}}{{\left( \frac{1}{3} \right)}^{3}}{{\left( \frac{2}{3} \right)}^{5}}\]
\[\therefore E\left( \text{3 success} \right)=100\times {}^{8}{{C}_{3}}{{\left( \frac{1}{3} \right)}^{3}}{{\left( \frac{2}{3} \right)}^{5}}\]

 Example 04

An unbiased coin is tossed six times. What is the probability that the tossed will result in:
1. Exactly two heads
2. At least five heads
3. At most two heads
4. Not greater than one head
5. Not less than five heads
6. At least one head

Solution:

Do it yourself.

 Example 05

The probability that an employee getting occupational disease is 20%. In a firm having five employees, what is the probability that:
1. None of the employee get the disease
2. Exactly two will get the disease
3. More than four will contract the disease

Solution:

Do it yourself.

 

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