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
bad risk.
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.
Thanks in advance! 🙂
Best Answer
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 $0.2\cdot 0.04 = 0.008.$
Similarly, $0.12\cdot 0.55=0.066$ and $0.3\cdot 0.25 = 0.075$ 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 $$\boxed{44.29\%}$$