Question

Suppose that x is a binomial random variable with n=5, p=. 3,and q=. 7.1. Write the binomial formula for this situation and listthe possible value of x. For each value of x calculate p(☓ =x)

Answer 1

`P(X = x) = C_(5,x).(0.3)^(x).(0.7)^(5-x)`

Possible values of x: Any from 0 to 5.

`P(X = 0) = C_(5,0).(0.3)^(0).(0.7)^(5-0) = 0.16807`

`P(X = 1) = C_(5,1).(0.3)^(1).(0.7)^(5-1) = 0.36015`

`P(X = 2) = C_(5,2).(0.3)^(2).(0.7)^(5-2) = 0.3087`

`P(X = 3) = C_(5,3).(0.3)^(3).(0.7)^(5-3) = 0.1323`

`P(X = 4) = C_(5,4).(0.3)^(4).(0.7)^(5-4) = 0.02835`

`P(X = 5) = C_(5,5).(0.3)^(5).(0.7)^(5-5) = 0.00243`

Explanation:

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

In which
`C_(n,x)` is the number of different combinations of x objects from a set of n elements, given by the following formula.

`C_(n,x) = (n!)/(x!(n-x)!)`

And p is the probability of X happening.

In this question:

`n = 5, p = 0.3, q = 1 - p = 0.7`

So

`P(X = x) = C_(5,x).(0.3)^(x).(0.7)^(5-x)`

Possible values of x: 5 trials, so any value from 0 to 5.

For each value of x calculate p(☓ =x)

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

`P(X = 0) = C_(5,0).(0.3)^(0).(0.7)^(5-0) = 0.16807`

`P(X = 1) = C_(5,1).(0.3)^(1).(0.7)^(5-1) = 0.36015`

`P(X = 2) = C_(5,2).(0.3)^(2).(0.7)^(5-2) = 0.3087`

`P(X = 3) = C_(5,3).(0.3)^(3).(0.7)^(5-3) = 0.1323`

`P(X = 4) = C_(5,4).(0.3)^(4).(0.7)^(5-4) = 0.02835`

`P(X = 5) = C_(5,5).(0.3)^(5).(0.7)^(5-5) = 0.00243`