(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? \( 70.67 \%, 35.33 \) of 50 points P(worker drives a car or takes public transportation to work) \( =0.8146 \) (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) \( =\square \) (Round to four decimal places as needed.)

To find the probability that a randomly selected worker primarily neither drives a car nor takes public transportation to work, we can use the complement of the probabilities from part (a). Given that the probability of driving a car or taking public transportation is 0.8146, we can compute the complementary probability. The probability that a worker neither drives a car nor takes public transportation can be calculated as follows: P(worker neither drives a car nor takes public transportation) = 1 - P(worker drives a car or takes public transportation) Substituting the value we found in part (a): P(worker neither drives a car nor takes public transportation) = 1 - 0.8146 = 0.1854 So the answer is: P(worker neither drives a car nor takes public transportation to work) = 0.1854 (rounded to four decimal places). Now you’re off to part (c)!