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 a warranty repair is 0.13, the probability that the car was manufactured by a domestic company is 0.66, and the probability that the new car needs a warranty repair and was manufactured by a domestic company is 0.047. 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?
0.13 (do not round.)
b. what is the probability that a new car selected at random needs a warranty repair and was manufactured by a domestic company?
0.047 (do not round.)
c. what is the probability that a new car selected at random needs a warranty repair or was manufactured by a domestic company?
(do not round.)

Explanation:

Part (a)

Step1: Identify given probability

The problem states the probability that a new car needs a warranty repair is given as \( 0.13 \).

Step1: Identify given probability

The problem states the probability that a new car needs a warranty repair and was manufactured by a domestic company is given as \( 0.047 \).

Step1: Recall the formula for \( P(A \cup B) \)

The formula for the probability of the union of two events \( A \) (needs warranty repair) and \( B \) (manufactured by domestic company) is \( P(A \cup B)=P(A)+P(B)-P(A \cap B) \).

Step2: Identify the values

We know that \( P(A) = 0.13 \) (probability of warranty repair), \( P(B)=0.66 \) (probability of domestic company), and \( P(A \cap B) = 0.047 \) (probability of both).

Step3: Substitute into the formula

\( P(A \cup B)=0.13 + 0.66- 0.047 \)
\( P(A \cup B)=0.743 \)

Answer:

\( 0.13 \)

Part (b)