A local car dealership currently has 35 used Company A, Company B, and Company C

Question

A local car dealership currently has 35 used Company A, Company B, and Company C vehicles that can be classifed as either cars or trucks. The following data are available. Complete parts a) through g) below 27 vehicles are ca 10 vehicles are Company A 15 vehicles are Company B 2 vehicles are both Company C and trucks 14 vehicles are both Company B and cars d) What is the probability that a randomly selected vehicle is a Company A truck? (Round to three decimal places as neoded.) (Round to three decimal places as needed.) ? (Round to three decimal places as needed.) e) What is the probability that a randomly selected vehicle is a Company C, given it is a car? ) What is the probability that a randomly selected vehicle is a truck, given it is a Company B? g) Construct a d tree for these events. Select the correct choice below and fill in the answer boxes within your choice. (Round to three decimal places as needed.) 1. P(truck and A) P(truck)2. P(track and B) l. P(A and car) 2. P(A and track) 3. P(B and car) 4. P(B and truck) MA) 3. Pttruck and C) 4. P(car and A) P(car) 5. P(car and B) 6. P(car and C) P(car) P(A) P(truck)- ? Click to select your answer(s).

Explanation / Answer

P(Truck)=8/35=0.2286

1.P(Truck & A)=P(Truck)*P(A|Truck)=0.2286*0.625=0.1429

2.P(Truck & B)=P(Truck)*P(B|Truck)=0.2286*0.125=0.02858

3.P(Truck & C)=P(Truck)*P(C|Truck)=0.2286*0.25=0.05715

P(Car)=27/35=0.7714

4. P(Car & A)=P(Car)*P(A|Car)=0.7714*0.1852=0.14285

5.P(Car & B)=P(Car)*P(B|Car)=0.7714*0.5185=0.39999

6.P(Car & C)=P(Car)*P(C|Car)=0.7714*0.2963=0.22856

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