A recent study found that 15.0% of passenger vehicles had defective tires and th
ID: 3176439 • Letter: A
Question
A recent study found that 15.0% of passenger vehicles had defective tires and that 12.0% had defective brakes. Assume further that given a passenger vehicle has defective tires, then the probability that it has defective brakes is 0.22. in what follows, let T be the event that a randomly chosen vehicle has defective tires and B be the event that it has defective brakes. Express the given information in probability notation. Are the events B and T independent? Explain your answer without performing any further calculations. Find P(B Intersection T). Find P(B union T). Given that a randomly chosen vehicle is found to have defective brakes, what is the probability that it has defective tires. Express your answer using probability notation.Explanation / Answer
a) P(T) = 0.15
P(B) = 0.12
P(B | T) = 0.22
b) Since, P(B | T) is not equal to P(B), B and T are not independent events
c) P(B | T) = P(B T) / P(T)
or, 0.22 = P(B T) / 0.15
or, P(B T) = 0.033
d) P(B U T) = P(B) + P(T) - P(B T)
= 0.15 + 0.12 - 0.033
= 0.237
e) P(T | B) = P(B T) / P(B)
= 0.033 / 0.12
= 0.275