(5.1,5.2)
According to the U.S. Census Bureau, the probability that a randomly selected worker primarily drives a car to work is 0.7637 . The probability that a randomly selected worker primarily takes public
transportation to work is 0.0509 . Complete parts (a) through (d).
(a) What is the probability that a randomly selected worker primarily drives a car or takes public transportation to work? of 50 points
P(worker drives a car or takes public transportation to work)
(Round to four decimal places as needed.)
(b) What is the probability that a randomly selected worker primarily neither drives a car nor takes public transportation to work?
P(worker neither drives a car nor takes public transportation to work)
(Round to four decimal places as needed.)

Ask by Mitchell Schwartz. in the United States

Mar 19,2025

Question

(5.1,5.2)
According to the U.S. Census Bureau, the probability that a randomly selected worker primarily drives a car to work is 0.7637 . The probability that a randomly selected worker primarily takes public
transportation to work is 0.0509 . Complete parts (a) through (d).
(a) What is the probability that a randomly selected worker primarily drives a car or takes public transportation to work? of 50 points
P(worker drives a car or takes public transportation to work)
(Round to four decimal places as needed.)
(b) What is the probability that a randomly selected worker primarily neither drives a car nor takes public transportation to work?
P(worker neither drives a car nor takes public transportation to work)
(Round to four decimal places as needed.)

Ask by Mitchell Schwartz. in the United States

Mar 19,2025

UpStudy AI · Accepted Answer

Let $$ p(\text{car}) = 0.7637 $$ and $$ p(\text{public transportation}) = 0.0509 $$.

**Step 1. Compute the probability of a worker driving a car or taking public transportation.**

Since these two events are mutually exclusive, we add the probabilities:

$$
p(\text{car or public transportation}) = p(\text{car}) + p(\text{public transportation}) = 0.7637 + 0.0509 = 0.8146
$$

**Step 2. Compute the probability of a worker primarily doing neither.**

This is the complement of the event in Step 1:

$$
p(\text{neither}) = 1 - p(\text{car or public transportation}) = 1 - 0.8146 = 0.1854
$$

Thus, the probability that a randomly selected worker primarily neither drives a car nor takes public transportation is

$$
\boxed{0.1854}
$$
The probability that a randomly selected worker primarily neither drives a car nor takes public transportation is 0.1854.