A recent study found that 15.0% of passenger vehicles had defective tires and th

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

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