The binomial distribution is one of the most useful probability distribution in statistic. It is a applicable in varieties of situations. Consider we have a sequence of n identical trials, where each trial can be result in one of two possibilities, known as success and failure, the experiment is called Bernoulli trial.

If we toss a coin, outcome will be "head" or "tail". In that one we can choose as "success" and other as "failure".

Binomial distribution is applicable to data where one of two mutually exclusive and independent outcomes are possible as a result of a single experiment. The experiment is a Bernoulli trial, if binomial distribution is applied. If we toss a coin, outcome will be "head" or "tail". In that one we can choose as "success" and other as "failure".

A random variable X is said to follow Binomial distribution with parameters n and p ( probability of success) if its probability density function is

f(r) = $^nC_r$ $p^r$ $q^{n-r}$

where r = 0, 1, 2, ...

**1)**p always lies between 0 and 1 (0< p<1)

**2)**p + q = 1 or q = 1 - p

The binomial probability formula is the result of getting exactly r successes in n number of trials.

The Binomial probability formula is given as follow:

P(r success in n trials) = $^nC_r$ $p^r$ $q^{n-r}$

**Where,**

p = the probability of success and

q = the probability of failure (or complement of the event)

n = Total number of trials

r = number of specific events we want to obtain

Also $^nC_r$ represents selection of r events from n, it can be written as: $^nC_r$ = $\frac{n!}{r!(n-r)!}$

**Lets have a look on some examples based on Binomial Probability:**

**Example 1:**Probability that a batsman scores a century in a cricket match is $\frac{2}{3}$. What is the probability that out of 5 matches, he may score century in:

**i)**exactly 2 matches

**ii)**none of the matches

**Solution:**

Here "success" is denoted by "scoring century"

Probability to scores a century in a cricket match is $\frac{2}{3}$.

Say, p = $\frac{2}{3}$

"Failure" is denoted by "not scoring century". We know that

q = 1 - p = 1 - $\frac{2}{3}$ = $\frac{1}{3}$.

Total number if matches n = 5.

Binomial probability formula is given by $^nC_r$ $p^r$ $q^{n-r}$

**i)**Find the probability that he scores century in exactly two matches.

That is r = 2.

P(scoring century in exactly 2 matches)

= $^5C_2$ $(\frac{2}{3})^2(\frac{1}{3})^{5-2}$

=$^5C_2$ $(\frac{2}{3})^2(\frac{1}{3})^3$

= 10 * ($\frac{4}{9}$)*($\frac{1}{27}$)

= $\frac{40}{243}$

P(scoring century in exactly 2 matches) = $\frac{40}{243}$

**ii)**We have to find the probability that he scores century in none of the matches.

That is r = 0

P(scoring century in none of the matches)

= $^5C_0$ $(\frac{2}{3})^0(\frac{1}{3})^{5-0}$

=$^5C_0$ $(\frac{2}{3})^0(\frac{1}{3})^5$

= 1 * 1 *($\frac{1}{243}$)

= $\frac{1}{243}$

P(scoring century in none of the matches) = $\frac{1}{243}$

**Example 2:**A basket contains 70 good apples and 30 are spoiled. Three apples are drawn at random from basket. What is the probability that of the 3 apples,

**i)**Exactly two good apples

**ii)**At least one is good

**iii)**At most two are good

**Solution:**

Here "success" is denoted by "good apples"

Given there are 70 good apples and 30 bad apples.

That is P(good apples) = p = $\frac{70}{100}$ = 0.70.

"Failure" is denoted by "getting bad apples". We know that

q = 1 - p = 1 - 0.70 = 0.30.

Number of apples selected n = 3.

According to binomial probability formula = $^nC_r$ $p^r$ $q^{n-r}$

**i)**Probability of getting exactly two good apples.

Here r = 2.

P(getting exactly 2 good apples)

= $^3C_2$ $(0.70)^2(0.30)^{3-2}$

= $^3C_2$ $(0.70)^2(0.30)^1$

= 3 * (0.49)*(0.30)

= 0.441

P(getting exactly 2 good apples) = 0.441

**ii)**Probability of getting at least one good apple.

That is r = 1, 2 or 3.

P(getting at least one good apple)

= $^3C_1$ $(0.70)^1(0.30)^{3-1}$ + $^3C_2$ $(0.70)^2(0.30)^{3-2}$ + $^3C_3$ $(0.70)^3(0.30)^{3-3}$

=$^3C_1$ $ (0.70)^1(0.30)^2$ + $^3C_2$ $(0.70)^2(0.30)^1$ + $^3C_3$ $(0.70)^3(0.30)^0$

= 3 * (0.70)*(0.09) + 3 * (0.49)*(0.30) + 3 * (0.343)*(1)

= 0.189 + 0.441+ 0.343

P(getting at least one good apple) = 0.973

**iii)**Probability of getting at most two good apples.

That is r = 0, 1 or 2

P(getting at most two good apples)

= $^3C_0$ $(0.70)^0(0.30)^{3-0}$ + $^3C_1$ $(0.70)^1(0.30)^{3-1}$ + $^3C_2$$(0.70)^2(0.30)^{3-2}$

= $^3C_0$ $(0.70)^0(0.30)^3$ + $^3C_1$ $(0.70)^1(0.30)^2$ + $^3C_2$ $(0.70)^2(0.30)^1$

= 1 * (1)*(0.027) + 3 * (0.70)*(0.09) + 3 * (0.49)*(0.30)

= 0.027 + 0.189 + 0.441

P(getting at most two good apples) = 0.657.

**Example 3:**Consider families with 4 children. What is the probability of having

**i)**two boys and two girls

**ii)**at least one boy

**iii)**no boys

**iv)**at most two boys

**Solution:**Consider p = "success" = "getting boys"

Probability of getting a boy and getting a girls is $\frac{1}{2}$.

i.e. p = $\frac{1}{2}$.

Also q = "Failure" = denoted by "getting girls".

=> q = 1 - p = 1 - $\frac{1}{2}$ =$\frac{1}{2}$.

Number of children n = 4.

Apply binomial probability formula, p(r success in n trials) = $^nC_r$ $p^r$ $q^{n-r}$

**i)**Probability of getting exactly two boys and two girls.

That is r = 2.

P(getting exactly 2 boys)

= $^4C_2$ $(\frac{1}{2})^2(\frac{1}{2})^{4-2}$

= $^4C_2$ $(\frac{1}{2})^2(\frac{1}{2})^2$

= 6 * $\frac{1}{16}$ = $\frac{6}{16}$ = $\frac{3}{8}$

P(getting exactly 2 boys and 2 girls) = $\frac{3}{8}$

**ii)**Probability of getting no boys.

That is r = 0.

P(getting no boys)

= $^4C_0$ $(\frac{1}{2})^0(\frac{1}{2})^{4-0}$

= $^4C_0$ $(\frac{1}{2})^0(\frac{1}{2})^4$

= 1 * 1 * $\frac{1}{16}$ = $\frac{1}{16}$

P(getting no boys) = $\frac{1}{16}$

**iii)**Probability of getting at most two boys.

That is r = 0, 1 or 2

P(getting at most two boys)

= $^4C_0$ $(\frac{1}{2})^0(\frac{1}{2})^{4-0}$ + $^4C^1$ $(\frac{1}{2})^1(\frac{1}{2})^{4-1}$ + $^4C_2$ $(\frac{1}{2})^2(\frac{1}{2})^{4-2}$

= $^4C_0$ $(\frac{1}{2})^0(\frac{1}{2})^4$ + $^4C_1$ 1.$(\frac{1}{2})^1(\frac{1}{2})^3$ + $^4C_2$ $(\frac{1}{2})^2(\frac{1}{2})^2$

= 1 * (1)*($\frac{1}{16}$) + 4 * ($\frac{1}{2}$)*($\frac{1}{8}$) + 6 * ($\frac{1}{4}$)*($\frac{1}{4}$)

= $\frac{1}{16}$ + $\frac{4}{16}$ + $\frac{6}{16}$

P(getting at most two boys) = $\frac{11}{16}$

**Try Yourself:**

**1)**A box contains 100 bulbs, 20 of which are defective. 10 of the bulbs are selected for inspection. Find the probability that

**i)**all 10 are defective

**ii)**all 10 are good

**iii)**at least one is defective

**iv)**at most three are defective.

**2)**A coin is tossed 16 times. What is the probability of obtaining

**i)**exactly five heads

**ii)**10 or more tails