Question

Suppose an automobile insurance company classifies a driver as good, average, or bad. Of all their insured drivers, 3.21% are classified good, 51.40% are average, and the rest are classified as bad. Suppose for the coming year, a good driver has a 6.00% chance of having an accident, and average driver has 18.56% chance of having an accident, and a bad driver has a 38.15% chance. If a driver had an accident in the past year, what is the probability that he or she was an average driver?

Answer 1

Given,

P(Good Driver) = 0.0321

P(Average Driver) = 0.5140

P(Bad Driver) = 1 - 0.0321 - 0.5140 = 0.4539

P(Accident | Good Driver) = 0.06

P(Accident | Average Driver) = 0.1856

P(Accident | Bad Driver) = 0.3815

Hence by Bayes theorem:

P(Average Driver | Accident)

$= \frac{0.5140*0.1856}{0.0321*0.06+0.5140*0.1856+0.4539*0.3815}$

= 0.3527