Suppose that of all used cars of a particular year, make, and model, 40% have ba

Question

Suppose that of all used cars of a particular year, make, and model, 40% have bad brakes. You are considering buying a used car of this year, make, and model. You take the car to a mechanic to have the brakes checked. The mechanic is not infallible: the chance that the mechanic will correctly report that a car with bad brakes has bad brakes is 40%, and the chance that the mechanic will erroneously report that a car with good brakes has bad brakes is 30%. Assuming that the car you take to the mechanic is selected "at random" from the population of cars of that year, make, and model, the chance that the car's brakes are good given that the mechanic says its brakes are good is?

Explanation / Answer

Given P( Bad Brakes) = P(BB) = 0.4. Therefore P(Good Brakes) = P(GB) = 1 - 0.4 = 0.6

LetP(MBB) = probability that the mechanic declares the brakes as bad = P(MBB) and if the brakes are good then P(MGB)

P(Mechanic declares brakes as bad, given that they are bad) = P(MBB/BB) = 0.4

P(Mechanic declares brakes as bad, given that they are good) = P(MBB/GB) = 0.3. Therefore P(MGB/GB) = 0.7

We need to find P( the brakes are good, given that the mechanic said they are good) = P(GB/MGB)

By Bayes Theorem: P(GB/MGB) = P(MGB/GB) * P(GB) / P(MGB)

We need P(MGB). We need to find P(MBB) = P(MBB/BB) * P(BB) + P(MBB/GB) * P(GB) = (0.4*0.4) + (0.3*0.6) = 0.16 + 0.18 = 0.34.

Therefore P(MGB) = 1 - 0.34 = 0.66

Therefore P(GB/MGB) = P(MGB/GB) * P(GB) / P(MGB) = 0.7 * 0.6/ 0.66 = 0.6364 = 63.64%

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