Question

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

Answer 1

`C(n,x)\ p^x\ q^(n-x)=(21)/(128)` when n=7, x=2,
`p=(1)/(2)`

Explanation:

Given : Information n=7, x=2,
`p=(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=(1)/(2)`

`q=1-p=1-(1)/(2)=(1)/(2)`

Substitute all values in the formula,

`=^7C_2\ ((1)/(2))^2\ ((1)/(2))^(7-2)`

`=(7!)/(2!(7-2)!)\ ((1)/(2))^2\ ((1)/(2))^(5)`

`=(7!)/(2!5!)\ ((1)/(2))^(2+5)`

`=(7* 6* 5!)/(2* 5!)\ ((1)/(2))^(7)`

`=21*(1)/(128)`

`=(21)/(128)`

Therefore,
`C(n,x)\ p^x\ q^(n-x)=(21)/(128)` when n=7, x=2,
`p=(1)/(2)`