Question

Suppose that 70% of Americans have dental insurance, 50% of Americans have vision insurance, and 30% of Americans have both. What is the probability that an American chosen at random:

5. Are the events an “American has dental insurance” and an “American has vision insurance” disjoint events?

6. Are the events an “American has dental insurance” and an “American has vision insurance” independent events?

7. If two Americans are chosen at random, what is the probability that both have dental insurance?

Answer 1

Let A denote the event that Americans have dental insurance and let B denote the event that Americans have vision insurance.

So P(A) = 0.7 , P(B) = 0.5 , P(A and B) = 0.3

So P(A or B) = P(A) + P(B) - P(A and B) = 0.7 + 0.5 - 0.3 = 0.9

5) The event A and B will be disjoint if P(A or B) = P(A) + P(B)

But, P(A) + P(B) = 0.7 + 0.5 = 1.2 which is not equal to P(A or B)

Hence A and B are not disjoint.

6) the event A and B are independent if P(A and B) = P(A)*P(B)

Here, P(A)*P(B) = 0.7*0.5 = 0.35 which is not equal to P(A and B)

Hence the event are not independent.

7) The probability that both Americans chosen has dental insurance is = 0.7*0.7 = 0.49