Question

A couple is thinking of buying a used Camry from a local dealer. In order to make an informed decision, they look up the records of Camrys in Consumer Reports, and find that 15% have faulty transmissions. To get more information on a particular Camry at the local dealer, the couple hires a mechanic who will make a quick assessment. The mechanic’s quick assessment isn’t always right—when examining faulty Camrys the mechanic has correctly pronounced 75% “bad”, and not quite as good a record in judging non-faulty Camrys, correctly deeming 95% “good”. What is your assessment of the chance that the Camry has a faulty transmission?

a)Before the mechanic looks at the Camry.

b)If the mechanic declares the Camry “bad.”

c)If the mechanic says the car is “good.”

Answer 1

Let F be the event when outcome is Faulty transmission or "bad" based on consumer report , and F" be the event when outcome is Non faulty transmission based on consumer report.

P(F) = 0.15 ,

So by compliment rule , P(F') = 1 - P(F) = 1 - 0.15 = 0.85

Let B shows the event that mechanic declares the car is bad and G shows the event that mechanic declares the car is good, So

Probability of mechanic declaring Faulty transmission as BAD , P(B|F) = 0.75

by compliment rule , P(G|F) = 1 - P(B|F) = 1 - 0.75 = 0.25

Probability of mechanic declaring Non Faulty transmission as Good, P(G|F') = 0.95

by compliment rule , P(B|F') = 1 - P(G|F') = 1 - 0.95 = 0.05

a)Chances of Faulty transmission , before the mechanic looks at the Camry :

Based on Camry consumer report , it is given that P(F) = 0.15

b)Chances of Faulty transmission , If the mechanic declares the Camry “bad.” (From Bayes theorem)

$P(F|B) =\frac{ P(B|F) P(F)}{P(B|F) P(F) + P(B|F') P(F')} =\frac{0.75 * 0.15}{(0.75 * 0.15) +(0.05 * 0.85)}$

$P(F|B) =\frac{0.1125}{0.1125 + 0.0425}= 0.726$

So, Chances of Faulty transmission , if the mechanic declares the Camry "bad" = 0.726 or 72.6%

c) Chances of Faulty transmission F , If the mechanic declares the Camry “good.”

$P(F|G) =\frac{P(G|F) P(F)}{(P(G|F)P(F) + P(G|F')P(F'))}$

$P(F|G) =\frac{0.25 * 0.15}{(0.25 * 0.15)(0.95 * 0.85))}=\frac{0.0375}{0.0375 + 0.8075}$

$P(F|G) == \frac{0.0375}{0.845} = 0.0444$

Hence , Chances of Faulty transmission , if the mechanic declares the Camry "Good" is 0.0444 or 4.44%.