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".

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, ...

The Binomial probability formula is given as follow:

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

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: $^nC_r$ = $\frac{n!}{r!(n-r)!}$

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 $^nC_r$ $p^r$ $q^{n-r}$

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) = $\frac{40}{243}$

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

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 = $^nC_r$ $p^r$ $q^{n-r}$

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

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

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.

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

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) = $\frac{3}{8}$

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

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) = $\frac{11}{16}$