ProbabilityHard
Question
Suppose that of all used cars of a particular year, 30% have bad brakes. You are considering buying a used car of that year. You take the car to a mechanic to have the brakes checked. The chance that the mechanic will give you wrong report is 20%. Assuming that the car you take to the mechanic is selected ″at random″ from the population of cars of that year. The chance that the car′s brakes are good given that the mechanic says its brakes are good is -
Options
A.28/31
B.29/31
C.16/31
D.15/31
Solution
Bad brake = 0.3
P(E2) = not bad brake = 0.7
mechanic gives correct report P(A1) = 0.8
good brake come → bad brake
given that mechanic says brakes are good
Bad brake Good brake
(0.3) × (0.2) (0.7) (1- 0.2)
Probabilty that brakes are good =
P(E2) = not bad brake = 0.7
mechanic gives correct report P(A1) = 0.8
good brake come → bad brake
given that mechanic says brakes are good
Bad brake Good brake
(0.3) × (0.2) (0.7) (1- 0.2)
Probabilty that brakes are good =
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