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) =

(round to four decimal places as needed.)

Explanation:

Step1: Identify the events

Let \(A\) be the event that a worker drives a car to work (\(P(A) = 0.7643\)), and \(B\) be the event that a worker takes public transportation (\(P(B)=0.0511\)). These two events are mutually - exclusive (a worker can't primarily drive a car and primarily take public transportation to work at the same time).

Step2: Apply the addition rule for mutually - exclusive events

For mutually - exclusive events \(A\) and \(B\), the probability \(P(A\cup B)\) is given by \(P(A\cup B)=P(A)+P(B)\).
Substitute \(P(A) = 0.7643\) and \(P(B)=0.0511\) into the formula:
\(P(A\cup B)=0.7643 + 0.0511\)

Step3: Calculate the result

\(P(A\cup B)=0.8154\)

Answer:

\(0.8154\)