Question

In a large group of Americans in 2001-2002, the probability of visiting a doctor's office in the past 12 months was 0.80.

(a) For two randomly chosen Americans in 2001-2002, what is the probability of at least one of them (either the first and not the second, or the second and not the first, or both) visiting a doctor that year? (Round your answer to two decimal places.)

(b) Show how to get the answer to part (a) by subtracting from 1 the probability that neither of them visits a doctor.

A) 1 − (0.20)(0.20)

B) 1 − (0.80)(0.20)

C) 1 − (2)(0.20)

D) 1 − (0.80)(0.80)

E) 1 − 0.90 /0.10

Answer 1

(a) The probability of one person not visiting a doctor's office in a year is 1 - 0.80 = 0.20. To find the probability that neither of the two people visit a doctor, you multiply these probabilities because they are independent events (the choices of one person do not affect the choices of the other):

Probability that neither of the two people visit a doctor = 0.20 * 0.20 = 0.04

Now, to find the probability of at least one of them visiting a doctor, you subtract the probability of neither visiting from 1:

Probability of at least one person visiting = 1 - 0.04 = 0.96

So, the answer is (A) 1 − (0.20)(0.20).

(b) This approach also uses the same logic. To find the probability of at least one of them visiting, you subtract from 1 the probability that neither of them visits a doctor, which is 0.04 as calculated above:

1 - 0.04 = 0.96

So, this approach corresponds to the answer (A) as well.