Question

Assume that a procedure yields a binomial distribution with a trial repeated n = 5 times. Use some form of technology to find the probability distribution given the probability p = 0.517 of success on a single trial. (Report answers accurate to 4 decimal places.) k P(X = k) 0 1 2 3 4 5

Answer 1

`P(X = 0) = 0.0263`

`P(X = 1) = 0.1407`

`P(X = 2) = 0.3012`

`P(X = 3) = 0.3224`

`P(X = 4) = 0.1725`

`P(X = 5) = 0.0369`

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 problem we have that:

`n = 5, p = 0.517`

Distribution

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

`P(X = 0) = C_(5,0).(0.517)^(0).(0.483)^(5) = 0.0263`

`P(X = 1) = C_(5,1).(0.517)^(1).(0.483)^(4) = 0.1407`

`P(X = 2) = C_(5,2).(0.517)^(2).(0.483)^(3) = 0.3012`

`P(X = 3) = C_(5,3).(0.517)^(3).(0.483)^(2) = 0.3224`

`P(X = 4) = C_(5,4).(0.517)^(4).(0.483)^(1) = 0.1725`

`P(X = 5) = C_(5,5).(0.517)^(5).(0.483)^(0) = 0.0369`