# 3 category car insurance probability

Mathematics Asked by mizerex on September 29, 2020

I couldn’t find any 3 category car insurance examples. I was able to deduce part a but I could use some help trying to figure out part b) of the following:

An insurance company believes that people can be classified into three groups: good risk,
average risk or bad risk. Their statistics show that the probabilities of good, average and bad
risk individuals being involved in an accident in any one year are $$0.04$$, $$0.12$$ and $$0.3$$ respectively.
Assume that $$20$$% of the population can be classified as good risk, $$55$$% as average risk and $$25$$% as

a. Find the proportion of policy holders having accidents in any one year.

b. Suppose that a new policy holder has an accident within a year of purchasing a policy. Find
the probability that the policy holder is an average risk.

For part a, I think it was just asking for the total probability of an accident which was P(A) = P(good)P(A|good) + P(avg)P(A|avg) + P(bad)P(A|bad), which is: $$.2(.04)+.55(.12)+.25(.3) = .149.$$

I think part b) involves Bayes’ theorem where P(avg|A) = (P(A|avg)P(avg))/P(A), and where the denominator is the answer from part a, and the numerator are given parts from the statement.

20% is $$0.2$$. Since the probability of any of them getting an accident is $$0.04$$, the total amount out of the total population is $$0.2cdot 0.04 = 0.008.$$

Similarly, $$0.12cdot 0.55=0.066$$ and $$0.3cdot 0.25 = 0.075$$ are the probabilities for average and bad risk people getting an accident in one year.

$$0.008+0.066+0.075 = 0.149.$$

This means $$boxed{14.9%}$$ had accidents in one year.

The average people contribute $$6.6%$$, so $$frac{6.6%}{14.9%} approx 0.4429.$$

Therefore, for b the answer is $$boxed{44.29%}$$

Correct answer by FruDe on September 29, 2020

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