Probability that a truck stopped at a roadblock will have faulty brake

1.	Probability that a truck stopped at a roadblock will have faulty brakes or badly worn tires are 0.23 and 0.24, respectively. Also, the probability is 0.38 that a truck stopped at the roadblock will have faulty brakes and/or badly working tires. What is the probability that a truck stopped at this roadblock will have faulty breaks as well as badly worn tires?
Answer» Let B be the event that a truck stopped at the roadblock will have faulty brakes and T be the event that it will have badly worn tires. We have P (B) = 0.23, P (T) = 0.24 and P (B∪T) = 0.38 and P (B ∪ T) = P (B) + P (T) – P (B ∩ T) So 0.38 = 0.23 + 0.24 – P (B ∩ T) ⇒ P (B ∩ T) = 0.23 + 0.24 – 0.38 = 0.09

Probability that a truck stopped at a roadblock will have faulty brakes or badly worn tires are 0.23 and 0.24, respectively. Also, the probability is 0.38 that a truck stopped at the roadblock will have faulty brakes and/or badly working tires. What is the probability that a truck stopped at this roadblock will have faulty breaks as well as badly worn tires?

Answer»

Let B be the event that a truck stopped at the roadblock will have faulty brakes and T be the event that it will have badly worn tires. We have P (B) = 0.23, P (T) = 0.24 and P (B∪T) = 0.38

and P (B ∪ T) = P (B) + P (T) – P (B ∩ T)

So 0.38 = 0.23 + 0.24 – P (B ∩ T)

⇒ P (B ∩ T) = 0.23 + 0.24 – 0.38 = 0.09

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