Question

Evaluate C_n.xP^xQn-x For the given n=7, x=2, p=1/2

Answer 1

Answer:

$C(n,x)\ p^x\ q^{n-x}=\frac{21}{128}$ when n=7, x=2, $p=\frac{1}{2}$

Step-by-step explanation:

Given : Information n=7, x=2, $p=\frac{1}{2}$

To find : Evaluate $C(n,x)\ p^x\ q^{n-x}$

Solution :

The formula given is a binomial distribution,

$C(n,x)\ p^x\ q^{n-x}$ can be written as  $^nC_x\ p^x\ q^{n-x}$

Here, n=7 x=2 and $p=\frac{1}{2}$

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

Substitute all values in the formula,

$=^7C_2\ (\frac{1}{2})^2\ (\frac{1}{2})^{7-2}$

$=\frac{7!}{2!(7-2)!}\ (\frac{1}{2})^2\ (\frac{1}{2})^{5}$

$=\frac{7!}{2!5!}\ (\frac{1}{2})^{2+5}$

$=\frac{7\times 6\times 5!}{2\times 5!}\ (\frac{1}{2})^{7}$

$=21\times\frac{1}{128}$

$=\frac{21}{128}$

Therefore, $C(n,x)\ p^x\ q^{n-x}=\frac{21}{128}$ when n=7, x=2, $p=\frac{1}{2}$