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) = 0.1846
(round to four decimal places as needed.)
(c) what is the probability that a randomly selected worker primarily does not drive a car to work?
p(worker does not drive a car to work) =
(round to four decimal places as needed.)

Explanation:

Step1: Use the complement rule

The complement rule states that \(P(\text{not }A)=1 - P(A)\). Here, \(A\) is the event that a worker drives a car to work. Given \(P(\text{worker drives a car to work}) = 0.7643\).

Step2: Calculate the probability

Using the formula \(P(\text{worker does not drive a car to work})=1 - P(\text{worker drives a car to work})\)

Substitute the value: \(1-0.7643 = 0.2357\)

Answer:

\(0.2357\)