Question

3 2 1. A recent study found that 10.0% of passenger vehicles had defective tyres and 13.0% had defective brakes. Assume further that given that a passenger vehicle has a defective tyres, the probability that it has defective brakes is 0.2. In what follows, let T be the event that a randomly chosen vehicle has defective tyres and B be the event that it has defective brakes. (a) Express the given information in probability notation. (b) Are the events B and T independent? Explain your answer without performing any further calculations. (c) Find P(BOT) (d) Find P(BUT). (e) Given that a randomly chosen vehicle is found to have defective brakes, what is the probability that it has defective tyres. Express your answer using probability notation. 2 2 2

Answer 1

a) P(T) = 0.10

P(B) = 0.13

P(B | T) = 0.2

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, 0.2 = P(B ∩ T) / 0.10

or, P(B ∩ T) = 0.02

d) P(B U T) = P(B) + P(T) - P(B ∩ T)

                   = 0.13 + 0.10 - 0.02

                   = 0.21

e) P(T | B) = P(B ∩ T) / P(B)

                  = 0.02 / 0.13

                  = 0.153

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