Question

According to the nonprofit group Zero Population Growth, 78% of the US population lives in urban areas. Scientists at Princeton University and the University of Wisconsin report that about 15% of all American adults care for ill relatives. Suppose 11 percent of adults living in urban areas care for ill relatives, answer the following questions.

a. Set up a contingency table for this problem.

b. Give an example of a simple event.

c. Give an example of a joint event.

d. What is the probability of randomly selecting an adult from the US population who lives in an urban area and does not care for an ill relative?

e. Suppose an adult cares for an ill relative, what is the probability that an adult lives in an urban area?

Answer 1

(a)

	Cares	Does not Care	Total
Urban	78% x 11% = 8.58%	78% - 8.58% = 69.42%	78%
Rural	15% - 8.58% = 6.42%	85% - 69.42% = 15.58%	100% - 78% = 22%
Total	15%	100% - 15% = 85%	100%

(b)

"Percentage of adults living in Urban area" is a simple event.

(c)

"Percentage of adults living in Urban area and caring for ill relatives" is a joint event.

(d)

Probability = 69.42% / 100% = 0.6942

(e)

Probability = 8.58% / 15% = 0.572