Question

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at bobs auto plaza there are currently 10 new cars, 6 used cars, 14 new trucks, and 11 used trucks. bob is going to choose one of these vehicles at random to be the deal of the month. what is the probability that the vehicle that bob chooses is used or is a car? do not round intermediate computations, and round your answer to the nearest hundredth.

Explanation:

Step1: Calculate total number of vehicles

Total vehicles = \(10 + 6+14 + 11=41\)

Step2: Calculate number of used or cars

Number of used vehicles = \(6 + 11=17\), number of new cars = \(10\). But used or cars: used (includes used cars and used trucks) and new cars. Wait, no, correct approach: number of used vehicles \(=6 + 11 = 17\), number of cars (new + used) \(=10+6=16\). But using formula \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\). Let \(A\) be the event of choosing a used vehicle, \(B\) be the event of choosing a car. \(n(A)=17\), \(n(B)=16\), \(n(A\cap B) = 6\) (used cars). \(P(A\cup B)=\frac{n(A)+n(B)-n(A\cap B)}{n(\text{total})}=\frac{17 + 16- 6}{41}=\frac{27}{41}\approx0.66\)

Answer:

\(0.66\)