Respuesta :
Answer:
1. 0.0417
2. 0.1375
3. no
Step-by-step explanation:
lets define:
A as the probability that a car needs a warranty repair.
B as probability taht the car is manufactured by a company that is based in country X
we have the
prob(A) = 0.08
prob(B) = 0.60
prob(A∩B) = 0.025
a. we are to calculate the probability taht a car manufactured in country X needs a warranty repair.
= p(AIB) =
[tex]P(\frac{AnB}{p(B)}) \\[/tex]
= 0.025/0.60
= 0.0417
2. Probability that car needs warranty repair given it was not manufactured in country X.
P(AIB')
= p(A∩B')/P(B')
= 0.08-0.025/1-0.60
= 0.055/0.40
= 0.1375
3. no, Need for warranty repair and location are not independent
The probability that the car needs warranty repair is 0.0417, the probability that the car needs warranty repair was not manufactured in company X is 0.1375 and no, there is no need for warranty repair and the location of the company manufacturing the car is not independent.
Given :
- Ratings are compiled concerning the performance of new cars during the first 90 days of use.
- Based on the data collected, the probability that the new car needs a warranty repair is .08.
- The probability that the car is manufactured by a company based in country X is .60.
- The probability that the new car needs a warranty repair and was manufactured by a company based in country X is .025.
a) The probability that the car needs warranty repair is given by:
[tex]\rm P(A|B) = \dfrac{ P(A\cap B)}{P(B)}[/tex]
[tex]\rm P(A|B) = \dfrac{0.025}{0.60}[/tex]
P(A|B) = 0.0417
b) The probability that the car needs warranty repair was not manufactured in company X is given by:
[tex]\rm P(A|B') = \dfrac{ P(A\cap B')}{P(B')}[/tex]
[tex]\rm P(A|B') = \dfrac{0.08-0.025}{1-0.60}[/tex]
P(A|B') = 0.1375
c) No, there is no need for warranty repair and the location of the company manufacturing the car is not independent.
For more information, refer to the link given below:
https://brainly.com/question/23044118