Each year, ratings are compiled concerning the performance of new cars during th

Question

Each year, ratings are compiled concerning the performance of new cars during the first 90 days of use. Suppose that the cars have been categorized according to whether the car needs warranty-related repair (yes or no) and the country in which the company manufacturing the car is based (United States or not United States). Based on the data collected, the probability that the new car needs warranty repair is 0.04, the probability that the car was manufactured by a U.S.-based company is 0.60, and the probability that the new car needs a warranty repair and was manufactured by a U.S.-based company is 0.025. Construct a contingency table or a Venn diagram to evaluate the probabilities of a warranty-related repair. What is the probability that a new car selected at random a. needs a warranty repair? b. needs a warranty repair and was manufactured by a U.S.- based company? c. needs a warranty repair or was manufactured by a U.S.- based company? d. needs a warranty repair or was not manufactured by a U.S.-based company?

Explanation / Answer

This is an exercise inconditional probability. For any two events A and B, where P(B) ? 0, you have the conditional probability:

P( A | B ) = P( A ? B ) / P( B ) = P( B | A) * P(A) / P(B)

the above is read as: the probability of A given B is equal to the probability of A and B divided by the probability of B.



In this question we know

P(US based) = 0.60
P(repair) = 0.04
P(repair AND US based) = 0.025

a) find P( repair | US based)
= P(repair AND US based) / P(US based)
= 0.025 / 0.60
= 0.04166667


b) find P( repair | not US Based )

we know from thelaw of total probabilitythat

P(repair) = P(repair AND US based) + P(repair AND not US Based)

0.04 = 0.025 + P(repair AND not US Based)

P(repair AND not US Based) = 0.015

so P( repair | not US Based )
= P(repair AND not US Based) / P(not US Based)
= 0.015 / 0.4
= 0.0375



c) for two events to be independent then P(A | B) = P(A) * P(B)

here we have:

P( repair | US based ) = 0.04166667
and
P(repair) * P(US Based) = 0.04 * 0.6 = 0.024

P( repair | US based ) ? P(repair) * P(US based)

and thus the events are not independent.

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