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?
P (worker drives a car or takes public transportation to work)
(Round to four decimal places as needed.)

Question

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?
P (worker drives a car or takes public transportation to work)

(Round to four decimal places as needed.)

UpStudy AI · Accepted Answer

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

2. Since these events are mutually exclusive (a worker cannot primarily use both methods simultaneously), we add the probabilities:
   $$
   P(\text{car or public transportation}) = P(\text{car}) + P(\text{public transportation})
   $$

3. Substitute the given values:
   $$
   P(\text{car or public transportation}) = 0.7637 + 0.0509 = 0.8146
   $$

Thus, the probability that a randomly selected worker primarily drives a car or takes public transportation to work is $$ 0.8146 $$.
The probability that a randomly selected worker primarily drives a car or takes public transportation to work is 0.8146.

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