Question

Attempt the given questions.

Q1. Let X have a binomial distribution with n = 9 and p = 1/5. Find P(X=4), P(X=6) and P(X=9).​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that X have a binomial distribution with

$\rm :\longmapsto\:n = 9$

$\rm :\longmapsto\:p = \dfrac{1}{5}$

We know,

$\rm :\longmapsto\: p+ q = 1$

$\rm :\longmapsto\: \dfrac{1}{5} + q = 1$

$\rm :\longmapsto\:q = 1 - \dfrac{1}{5}$

$\rm :\longmapsto\:q = \dfrac{5 - 1}{5}$

$\rm :\longmapsto\:q = \dfrac{4}{5}$

Also,

Binomial Distribution is given by

$\rm :\longmapsto\:P(r) = \: ^nC_r \: {p}^{r} {q}^{n - r}$

where,

• n = number of independent trials

• p = probability of success

• q = probability of failure

• r = random variable

Consider,

$\bf :\longmapsto\:P(4)$

$\rm :\ = \: \: ^nC_4 \: {p}^{4} {q}^{n - 4}$

On substituting the values of n, p and q, we get

$\rm \: = \: \: ^9C_4 \: { \bigg(\dfrac{1}{5} \bigg)}^{4} { \bigg(\dfrac{4}{5} \bigg)}^{9 - 4}$

$\rm \: = \: \: \dfrac{9!}{4! \: 5!} \: { \bigg(\dfrac{1}{5} \bigg)}^{4} { \bigg(\dfrac{4}{5} \bigg)}^{5}$

$\rm \: = \: \: \dfrac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} \times \dfrac{ {4}^{5} }{ {5}^{9} }$

$\rm \: = \: \: 126 \times \dfrac{ {4}^{5} }{ {5}^{9} }$

$\bf\implies \:P(4) = \dfrac{126 \times {4}^{5} }{ {5}^{9} }$

Consider,

$\bf :\longmapsto\:P(6)$

$\rm :\ = \: \: ^nC_6 \: {p}^{6} {q}^{n - 6}$

On substituting the values of n, p and q, we get

$\rm \: = \: \: ^9C_6 \: { \bigg(\dfrac{1}{5} \bigg)}^{6} { \bigg(\dfrac{4}{5} \bigg)}^{9 - 6}$

$\rm \: = \: \: \dfrac{9!}{6! \: 3!} \: { \bigg(\dfrac{1}{5} \bigg)}^{6} { \bigg(\dfrac{4}{5} \bigg)}^{3}$

$\rm \: = \: \: \dfrac{9 \times 8 \times 7 }{ 3 \times 2 \times 1} \times \dfrac{ {4}^{3} }{ {5}^{9} }$

$\rm \: = \: \: 84 \times \dfrac{ {4}^{3} }{ {5}^{9} }$

$\bf\implies \:P(6) = \dfrac{84\times {4}^{3} }{ {5}^{9} }$

Consider,

$\bf :\longmapsto\:P(6)$

$\rm :\ = \: \: ^nC_9 \: {p}^{9} {q}^{n - 9}$

On substituting the values of n, p and q, we get

$\rm \: = \: \: ^9C_9 \: { \bigg(\dfrac{1}{5} \bigg)}^{9} { \bigg(\dfrac{4}{5} \bigg)}^{9 - 9}$

$\rm \: = \: \: \dfrac{9!}{9! \: 0!} \: { \bigg(\dfrac{1}{5} \bigg)}^{9} { \bigg(\dfrac{4}{5} \bigg)}^{0}$

$\rm \: = \: \: \dfrac{1}{ {5}^{9} }$

$\bf\implies \:P(9) = \dfrac{1}{ {5}^{9} }$

Additional Information :-

In Binomial Distribution,

• Mean = np

• Variance = npq

• Mean > Variance