Question

By some estimates, 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. (a) What is the probability that exactly 3 of the 15 sampled have no health insurance? (b) What is the probability that at most 1 of those sampled have no health insurance?

Answer 1

Answer:

a) 25.01% probability that exactly 3 of the 15 sampled have no health insurance.

b) 16.71% probability that at most 1 of those sampled have no health insurance

Step-by-step explanation:

For each person in the sample, there are only two possible outcomes. Either they have no health insurance, or they have. The probability of a person having health insurance is independent of any other person. So we use the binomial probability distriution 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.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

By some estimates, twenty-percent (20%) of Americans have no health insurance.

This means that $p = 0.2$

Randomly sample n = 15 Americans

This means that $n = 15$.

(a) What is the probability that exactly 3 of the 15 sampled have no health insurance?

This is P(X = 3).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 3) = C_{15,3}.(0.2)^{3}.(0.8)^{12} = 0.2501$

25.01% probability that exactly 3 of the 15 sampled have no health insurance.

(b) What is the probability that at most 1 of those sampled have no health insurance?

$P(X \leq 1) = P(X = 0) + P(X = 1)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{15,0}.(0.2)^{0}.(0.8)^{15} = 0.0352$

$P(X = 1) = C_{15,1}.(0.2)^{1}.(0.8)^{14} = 0.1319$

$P(X \leq 1) = P(X = 0) + P(X = 1) = 0.0352 + 0.1319 = 0.1671$

16.71% probability that at most 1 of those sampled have no health insurance