\begin{tabular}{l}$$ 10(5.1,5.2) $$ \ 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 publio \ 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) $$ =0.8146 $$ \ (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? 5.2 .38 of 50 points \ P(worker neither drives a car nor takes public transportation to work) $$ =0.1854 $$ \ (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) $$ =0.2363 $$ \ (Round to four decimal places as needed.) \ (d) Can the probability that a randomly selected worker primarily walks to work equal $$ 0.16 ? $$ Why or why not? \ \hline\end{tabular}

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UpStudy AI · Accepted Answer

**(a)**  
We are given the probability that a worker drives a car is  
$$
P(\text{car}) = 0.7637,
$$  
and the probability that a worker takes public transportation is  
$$
P(\text{public transportation}) = 0.0509.
$$  
Since these events are mutually exclusive, the probability that a worker does one or the other is  
$$
P(\text{car or public transportation}) = 0.7637 + 0.0509 = 0.8146.
$$

**(b)**  
The probability of the complementary event (neither drives a car nor takes public transportation) is  
$$
P(\text{neither}) = 1 - P(\text{car or public transportation}) = 1 - 0.8146 = 0.1854.
$$

**(c)**  
The probability that a worker does not drive a car is the complement of the probability that a worker drives a car:  
$$
P(\text{not driving}) = 1 - P(\text{car}) = 1 - 0.7637 = 0.2363.
$$

**(d)**  
Walking to work is one mode of transportation included in the $$0.1854$$ probability (the probability that a worker neither drives a car nor takes public transportation). Therefore, the probability that a worker primarily walks to work must satisfy  
$$
0 \leq P(\text{walk}) \leq 0.1854.
$$  
Since $$0.16 \leq 0.1854$$, it is possible for  
$$
P(\text{walk}) = 0.16,
$$  
provided the remaining probability of $$0.1854 - 0.16 = 0.0254$$ is distributed among other modes (such as biking or other means). Thus, the probability that a worker primarily walks to work can equal $$0.16$$.
**(a)** The probability that a worker drives a car or takes public transportation is $$0.8146$$. **(b)** The probability that a worker neither drives a car nor takes public transportation is $$0.1854$$. **(c)** The probability that a worker does not drive a car to work is $$0.2363$$. **(d)** Yes, the probability that a worker primarily walks to work can be $$0.16$$ because it is within the range of the remaining probability after accounting for driving and public transportation.

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