Answer:
[tex]C(n,x)\ p^x\ q^{n-x}=\frac{21}{128}[/tex] when n=7, x=2, [tex]p=\frac{1}{2}[/tex]
Step-by-step explanation:
Given : Information n=7, x=2, [tex]p=\frac{1}{2}[/tex]
To find : Evaluate [tex]C(n,x)\ p^x\ q^{n-x}[/tex]
Solution :
The formula given is a binomial distribution,
[tex]C(n,x)\ p^x\ q^{n-x}[/tex] can be written as [tex]^nC_x\ p^x\ q^{n-x}[/tex]
Here, n=7 x=2 and [tex]p=\frac{1}{2}[/tex]
[tex]q=1-p=1-\frac{1}{2}=\frac{1}{2}[/tex]
Substitute all values in the formula,
[tex]=^7C_2\ (\frac{1}{2})^2\ (\frac{1}{2})^{7-2}[/tex]
[tex]=\frac{7!}{2!(7-2)!}\ (\frac{1}{2})^2\ (\frac{1}{2})^{5}[/tex]
[tex]=\frac{7!}{2!5!}\ (\frac{1}{2})^{2+5}[/tex]
[tex]=\frac{7\times 6\times 5!}{2\times 5!}\ (\frac{1}{2})^{7}[/tex]
[tex]=21\times\frac{1}{128}[/tex]
[tex]=\frac{21}{128}[/tex]
Therefore, [tex]C(n,x)\ p^x\ q^{n-x}=\frac{21}{128}[/tex] when n=7, x=2, [tex]p=\frac{1}{2}[/tex]
