Of the travelers arriving at a gas station, 60% drive compact cars, 30% drive se

Question

Of the travelers arriving at a gas station, 60% drive compact cars, 30% drive semi-trucks, and the remainder drive minivans or SUVs. Of those driving compact cars, 50% are over 100 miles from home, whereas 60% of those driving semis and 90% of those driving minivans and SUVs are over 100 miles from home. Suppose we randomly select a driver arriving at this gas station. What is the probability that the person. (a) is over 100 miles from home? (b) is over 100 miles from home driving a semi? (c) is driving a semi, given that the person is over 100 miles from home? (d) is 100 miles from home, given that the person is driving a minivan or an SUV?

Explanation / Answer

(a) Probability = 57/100 = 0.57

(b) Probability = 30/100 = 0.30

(c) Probability = (30/100)/(57/100) = 30/57 = 0.526

(d) Probability = [(18 + 9)/100] / [(30 + 10)/100] = 27/40 = 0.675

Compact ST SUV Total > 100 miles 50% of 60 = 30 60% of 30 = 18 90% of 10 = 9 57 100 miles 60 - 30 = 30 30 - 18 = 12 10 - 9 = 1 43 Total 60 30 10 100

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