Question

A country conducts a study on new cars within the first 90 days of use. The cars have been categorized according to whether the car needs a​ warranty-based repair​ (yes or​ no) and the​ car's origin​ (domestic or​ foreign). Based on the data​ collected, the probability that the new car needs warranty repair is 0.09, the probability that the car was manufactured by a domestic company is 0.73, and the probability that the new car needs a warranty repair and was manufactured by a domestic company is 0.037. Construct a contingency table to evaluate the probabilities of a​ warranty-related repair. Complete parts​ (a) through​ (d).

a. What is the probability that a new car selected at random needs a warranty​ repair?

b. What is the probability that a new car selected at random needs a warranty repair and was manufactured by a domestic​ company?

c. What is the probability that a new car selected at random needs a warranty repair or was manufactured by a domestic​ company?

d. What is the probability that a new car selected at random needs a warranty repair or was made by a foreign​ company?

Answer 1

a)

probability that a new car selected at random needs a warranty​ repair =0.09

b) probability that a new car selected at random needs a warranty repair and was manufactured by a domestic​ company =0.037

c)probability that a new car selected at random needs a warranty repair or was manufactured by a domestic​ company

=P(A u B)=P(A)+P(B)-P(A n B) =0.09+0.73-0.037 =0.783

d)

probability that a new car selected at random needs a warranty repair and was made by a foreign​ company

=P(need warranty repair)-P( needs a warranty repair and was manufactured by a domestic​ company)=0.09-0.037=0.053

P( made by a foreign​ company ) =1-P(made by domestic​ company ) =1-0.73 =0.27

hence  probability that a new car selected at random needs a warranty repair or was made by a foreign​ company

=0.09+0.27-0.053 =0.307