Question

Please help to solve thos question..

Twenty-percent (20%) of Americans have no health insurance. Randomly sample n = 15 Americans. Let X denote the number in the sample with no health insurance. What is the probability that 3 of them have insurance? ​

Answer 1

0% probability that 3 of them have insurance

Explanation:

For each American, there are only two possible outcomes. Either they have insurance, or they do not. So we use the binomial probability distribution to solve this question.

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.

Twenty-percent (20%) of Americans have no health insurance.

This means that 80% have insurance, so
`p = 0.8`

Randomly sample n = 15 Americans.

This means that
`n = 15`

What is the probability that 3 of them have insurance? ​

This is
`P(X = 3)`

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

`P(X = 3) = C_(15,3).(0.8)^(3).(0.2)^(12) = 0`

0% probability that 3 of them have insurance