Question

according to the u s census bureau, the probability that a randomly selected worker primarily drives a car to work is 0 7643 the probability that a randomly selected worker primarily takes public transportation to work is 0 0511 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?
p(worker drives a car or takes public transportation to work) = 0 8154
(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.)

Explanation:

Step1: Use the complement rule

The complement rule states that \(P(\text{not }A)=1 - P(A)\). Let \(A\) be the event that a worker primarily drives a car or takes public transportation to work. We know from part (a) that \(P(A) = 0.8154\).

Step2: Calculate the probability

We want to find \(P(\text{worker neither drives a car nor takes public transportation to work})\). Using the complement rule:

$$P = 1-0.8154$$

$$P=0.1846$$

Answer:

\(0.1846\)