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) = 0.2357 (round to four decimal places as needed.) (d) can the probability that a randomly selected worker primarily walks to work equal 0.14? why or why not? be possible because the probability a worker primarily drives, walks, or takes public transportation would be

Explanation:

Step1: Calculate the sum of probabilities

The probability that a worker drives a car is \(P(\text{car}) = 0.7643\), and the probability that a worker takes public transportation is \(P(\text{public})=0.0511\). If we assume the probability of walking is \(P(\text{walk}) = 0.14\), then the sum of these probabilities is \(P = 0.7643+0.0511 + 0.14\).
$$P=0.7643+0.0511+0.14=0.9554$$

Step2: Check the probability rule

The probability of any event or combination of mutually - exclusive events (in this case, driving, taking public transport, and walking are assumed to be mutually - exclusive ways of commuting) must be less than or equal to \(1\).

Answer:

No, because the sum of the probabilities (if \(P(\text{walk}) = 0.14\)) would be \(0.9554\) (less than \(1\)), but in reality, there are other commuting methods (biking, etc.). However, if we consider only these three methods (driving, public transport, walking) as the only possible mutually - exclusive commuting methods, the sum \(0.7643 + 0.0511+0.14=0.9554<1\) which violates the rule that the sum of probabilities of all mutually - exclusive and exhaustive events is \(1\). So, if we assume that there are no other commuting methods (which is not true in the real - world context, but based on the given data and the assumption of only these three methods being considered), it is not possible because the sum of probabilities of all mutually - exclusive commuting methods (driving, public transport, walking) would be \(0.7643 + 0.0511+0.14 = 0.9554<1\) (a probability value must be between \(0\) and \(1\) for a single event, and for a set of mutually - exclusive and exhaustive events, their sum must be \(1\)).