3.015) The OWL Insurance Company sells automobile insurance policies and has bee

Question

3.015) The OWL Insurance Company sells automobile insurance policies and has been tracking the accident history of its policyholders, all of whom are long-term customers of this insurance company. After examining accidents in 2014 and 2015, they determine that among the policyholders who had zero accidents in 2014, the proportion of policyholders who had at least one accident in 2015 is 0.20. The probability that a policyholder has zero accidents in 2014 and zero accidents in 2015 is 0.60. Among the policyholders who had at least one accident in 2014, 70% had at least one accident in 2015.

A policyholder is picked at random from the group of policyholders who had zero accidents in 2015. What is the probability this policyholder had zero accidents in 2014?

Answer is .889. not sure how to get there.

Explanation / Answer

Here the given data:

Let say A = Zero accident in 2014 and B = Zero accident in 2015

A' = At least one accident in 2014 and B' = at least one accident in 2015

Pr(B' l A) = 0.20

Pr( A B) = 0.60

Pr(B' l A' ) = 0.70

Now we have to find,

Pr(A l B)

by applying bayes rule and conditional probability rule

Pr( A B) = 0.60 = P(A) * P(B l A) = P(B) * P(A l B)

Pr(B' l A) = 0.20

P(B l A) = 1 - 0.20 = 0.80

P(A) * P(B l A) = 0.60

P(A) = 0.60/0.80 = 0.75

P(B l A') = 1 - P(B' l A') = 1 - 0.70 = 0.30

so P (A l B) = P(B l A) * P(A) / [P(B l A) * P(A) + P(B l A') * P(A')]

P(A l B) = 0.80 * 0.75 / (0.80 * 0.75 + 0.30 * 0.25)

= 8/9

= 0.889

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